Find the Jordan forms of a matrix from just the ranks of its eigenspaces
Let $C$ be a $10times 10$ matrix whose characteristic polynomial is $(t+2)^5(t-3)^5$. Suppose that $rank((C+2I)^2)=6$ and $rank((C-3I))=8$. What are the possible Jordan forms of $C$?
This is the first problem in which I have ever encountered Jordan forms, so when it says to find all possible Jordan forms of $C$, I take that to mean finding every combination of possible Jordan blocks. Since $rank((C-3I))=8$, I know by rank nullity theorem that the dimension of the eigenspace of $lambda=3$ is $2$. This tells me that there will be two Jordan blocks for $lambda=3$, but I don't know how to determine the respective sizes because I don't know what the ranks of higher powers of $(C-3I)$ are. Additionally, I don't know how to glean any information from $rank(C+2I)^2=6$.
linear-algebra matrix-rank jordan-normal-form
asked Dec 4 '18 at 21:23
Ryan GreylingRyan Greyling
1827
add a comment |
Let $C$ be a $10times 10$ matrix whose characteristic polynomial is $(t+2)^5(t-3)^5$. Suppose that $rank((C+2I)^2)=6$ and $rank((C-3I))=8$. What are the possible Jordan forms of $C$?
This is the first problem in which I have ever encountered Jordan forms, so when it says to find all possible Jordan forms of $C$, I take that to mean finding every combination of possible Jordan blocks. Since $rank((C-3I))=8$, I know by rank nullity theorem that the dimension of the eigenspace of $lambda=3$ is $2$. This tells me that there will be two Jordan blocks for $lambda=3$, but I don't know how to determine the respective sizes because I don't know what the ranks of higher powers of $(C-3I)$ are. Additionally, I don't know how to glean any information from $rank(C+2I)^2=6$.
linear-algebra matrix-rank jordan-normal-form
asked Dec 4 '18 at 21:23
Ryan GreylingRyan Greyling
1827
So you have that for $lambda =3$ we have 2 Jordan blocks. The question doesn't ask which ones exactly. Instead it asks which blocks they might be. Can you find the 2 possible configurations?
– I like Serena
Dec 4 '18 at 23:37
Suppose $C$ is already in Jordan normal form. Now pick some configuration of the blocks, say, 1 big block for $lambda=-2$. What will $(C+2I)^2$ look like? And what is its rank?
– I like Serena
Dec 4 '18 at 23:37
Hint: the number of Jordan blocks of an eigenvalue is equal to its geometric multiplicity.
– amd
Dec 5 '18 at 3:47
add a comment |
Let $C$ be a $10times 10$ matrix whose characteristic polynomial is $(t+2)^5(t-3)^5$. Suppose that $rank((C+2I)^2)=6$ and $rank((C-3I))=8$. What are the possible Jordan forms of $C$?
This is the first problem in which I have ever encountered Jordan forms, so when it says to find all possible Jordan forms of $C$, I take that to mean finding every combination of possible Jordan blocks. Since $rank((C-3I))=8$, I know by rank nullity theorem that the dimension of the eigenspace of $lambda=3$ is $2$. This tells me that there will be two Jordan blocks for $lambda=3$, but I don't know how to determine the respective sizes because I don't know what the ranks of higher powers of $(C-3I)$ are. Additionally, I don't know how to glean any information from $rank(C+2I)^2=6$.
linear-algebra matrix-rank jordan-normal-form
asked Dec 4 '18 at 21:23
Ryan GreylingRyan Greyling
1827
Let $C$ be a $10times 10$ matrix whose characteristic polynomial is $(t+2)^5(t-3)^5$. Suppose that $rank((C+2I)^2)=6$ and $rank((C-3I))=8$. What are the possible Jordan forms of $C$?
This is the first problem in which I have ever encountered Jordan forms, so when it says to find all possible Jordan forms of $C$, I take that to mean finding every combination of possible Jordan blocks. Since $rank((C-3I))=8$, I know by rank nullity theorem that the dimension of the eigenspace of $lambda=3$ is $2$. This tells me that there will be two Jordan blocks for $lambda=3$, but I don't know how to determine the respective sizes because I don't know what the ranks of higher powers of $(C-3I)$ are. Additionally, I don't know how to glean any information from $rank(C+2I)^2=6$.
linear-algebra matrix-rank jordan-normal-form
linear-algebra matrix-rank jordan-normal-form
asked Dec 4 '18 at 21:23
Ryan GreylingRyan Greyling
1827
asked Dec 4 '18 at 21:23
Ryan GreylingRyan Greyling
1827
asked Dec 4 '18 at 21:23
Ryan GreylingRyan Greyling
1827
asked Dec 4 '18 at 21:23
Ryan GreylingRyan Greyling
1827
asked Dec 4 '18 at 21:23
Ryan GreylingRyan Greyling
1827
1827
So you have that for $lambda =3$ we have 2 Jordan blocks. The question doesn't ask which ones exactly. Instead it asks which blocks they might be. Can you find the 2 possible configurations?
– I like Serena
Dec 4 '18 at 23:37
Suppose $C$ is already in Jordan normal form. Now pick some configuration of the blocks, say, 1 big block for $lambda=-2$. What will $(C+2I)^2$ look like? And what is its rank?
– I like Serena
Dec 4 '18 at 23:37
Hint: the number of Jordan blocks of an eigenvalue is equal to its geometric multiplicity.
– amd
Dec 5 '18 at 3:47
add a comment |
So you have that for $lambda =3$ we have 2 Jordan blocks. The question doesn't ask which ones exactly. Instead it asks which blocks they might be. Can you find the 2 possible configurations?
– I like Serena
Dec 4 '18 at 23:37
Suppose $C$ is already in Jordan normal form. Now pick some configuration of the blocks, say, 1 big block for $lambda=-2$. What will $(C+2I)^2$ look like? And what is its rank?
– I like Serena
Dec 4 '18 at 23:37
Hint: the number of Jordan blocks of an eigenvalue is equal to its geometric multiplicity.
– amd
Dec 5 '18 at 3:47
So you have that for $lambda =3$ we have 2 Jordan blocks. The question doesn't ask which ones exactly. Instead it asks which blocks they might be. Can you find the 2 possible configurations?
– I like Serena
Dec 4 '18 at 23:37
So you have that for $lambda =3$ we have 2 Jordan blocks. The question doesn't ask which ones exactly. Instead it asks which blocks they might be. Can you find the 2 possible configurations?
– I like Serena
Dec 4 '18 at 23:37
Suppose $C$ is already in Jordan normal form. Now pick some configuration of the blocks, say, 1 big block for $lambda=-2$. What will $(C+2I)^2$ look like? And what is its rank?
– I like Serena
Dec 4 '18 at 23:37
Suppose $C$ is already in Jordan normal form. Now pick some configuration of the blocks, say, 1 big block for $lambda=-2$. What will $(C+2I)^2$ look like? And what is its rank?
– I like Serena
Dec 4 '18 at 23:37
Hint: the number of Jordan blocks of an eigenvalue is equal to its geometric multiplicity.
– amd
Dec 5 '18 at 3:47
Hint: the number of Jordan blocks of an eigenvalue is equal to its geometric multiplicity.
– amd
Dec 5 '18 at 3:47
add a comment |
1 Answer
1
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Throughout the post, I use $J(c;m)$ to denote the Jordan block with diagonal entries $c$ of size $mtimes m$.
According to the char. polynomial, the Jordan form $J$ of $C$ has five $-2$ and five $3$ on the diagonal. For Jordan blocks with diagonal entries $3$, yes you have 2 blocks. For Jordan blocks with diag. entries $-2$, easy to see that $DeclareMathOperatorrank{rank} rank (J(-2;m)+2I) = m-1$ and $rank ((J(-2;m)+2I)^2) = m-2$. Restricting to the blocks with diagonal entries $-2$, the total dimension is $5$ by the char. polynomial, so $rank(J(-2)^2)=1$, then there exists at least $1$ block $J(-2;3)$, and the rest of the blocks could be 2 $J(-2;1)$ or 1 $J(-2;2)$.
Conclusion: up to permutation, the possible Jordan forms are $DeclareMathOperatordiag{diag} diag(J(-2;3), J(-2;2), J(3;1), J(3;4))$, $diag (J(-2;3), J(-2;2), J(3;2), J(3;3))$, $diag(J(-2;3), -2, -2, 3, J(3;4))$ and $diag (J(-2;3), -2, -2, J(3;2), J(3;3))$.
UPDATE
You could prove that
$$
rankbegin{bmatrix} A & O \O & Bend{bmatrix} = rank A + rank B
$$
where the LHS is a block diagonal matrix.
Then, for example,
$$
J = begin{bmatrix} -2 & 1 &&&\ &-2&&&\ &&3 & * & * \ &&&3 &* \ &&&&3end{bmatrix} = mathrm {diag}(J(-2), J(3)),
$$
then
$$
(J+2I)^2= begin{bmatrix} 0 & 0 &&&\ &0&&&\ &&25 & * & * \ &&&25 &* \ &&&&25end{bmatrix} = mathrm {diag}((J(-2)+2I)^2, (J(3)+2I)^2),
$$
and $rank((J+2I)^2) = 3$, so
$$
rank((J(-2)+2I)^2) = 3 - rank((J(3)+2I)^2) = 3-3 = 0.
$$
answered Dec 5 '18 at 2:19
xbhxbh
5,7651522
What does $J(-2)$ mean? Also, since the dimension for diagonal entries $-2$ is $5$, how does this tell you that $rank(J(-2)^2)=1$? I know that this must involve the fact that $rank(C+2I)^2=6$ but I don't see how.
– Ryan Greyling
Dec 5 '18 at 4:42
@RyanGreyling $J(-2)$ is the biggest block with the diagonal entries $-2$. In this question, it is a upper triangular matrix with all diagonal entries $-2$. Example if the Jordan blocks are $J(3;4), J(3;5)$, then the $J(3) = mathrm {diag}(J(3;4), J(3;5))$.
– xbh
Dec 5 '18 at 4:47
@RyanGreyling For the 2nd question, note that for the Jordan form $J =mathrm {diag}(J(-2), J(3))$, $J+2I = mathrm {diag}(J(0), J(5))$, so the rank equals $mathrm {rank}(J(0)) + mathrm {rank}(J(5))$, where $J(5)$ is of full rank, i.e. 5.
– xbh
Dec 5 '18 at 4:51
Okay, I got it. The way I understand it uses the same logic in your explanation but with notation that makes more sense for me personally. Since $rank(C+2I)^2=6$, by rank-nullity theorem the dimension of the solution space for $(C+2I)^2vec{x}=0$ is $4$. $4<5$, the algebraic multiplicity for $lambda=-2$, so when drawing the box diagram for the Jordan form, there will be at least one column of $3$ boxes (one row above and one below). This means one of the Jordan blocks is of size $3$.
– Ryan Greyling
Dec 5 '18 at 19:03
add a comment |
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Throughout the post, I use $J(c;m)$ to denote the Jordan block with diagonal entries $c$ of size $mtimes m$.
According to the char. polynomial, the Jordan form $J$ of $C$ has five $-2$ and five $3$ on the diagonal. For Jordan blocks with diagonal entries $3$, yes you have 2 blocks. For Jordan blocks with diag. entries $-2$, easy to see that $DeclareMathOperatorrank{rank} rank (J(-2;m)+2I) = m-1$ and $rank ((J(-2;m)+2I)^2) = m-2$. Restricting to the blocks with diagonal entries $-2$, the total dimension is $5$ by the char. polynomial, so $rank(J(-2)^2)=1$, then there exists at least $1$ block $J(-2;3)$, and the rest of the blocks could be 2 $J(-2;1)$ or 1 $J(-2;2)$.
Conclusion: up to permutation, the possible Jordan forms are $DeclareMathOperatordiag{diag} diag(J(-2;3), J(-2;2), J(3;1), J(3;4))$, $diag (J(-2;3), J(-2;2), J(3;2), J(3;3))$, $diag(J(-2;3), -2, -2, 3, J(3;4))$ and $diag (J(-2;3), -2, -2, J(3;2), J(3;3))$.
UPDATE
You could prove that
$$
rankbegin{bmatrix} A & O \O & Bend{bmatrix} = rank A + rank B
$$
where the LHS is a block diagonal matrix.
Then, for example,
$$
J = begin{bmatrix} -2 & 1 &&&\ &-2&&&\ &&3 & * & * \ &&&3 &* \ &&&&3end{bmatrix} = mathrm {diag}(J(-2), J(3)),
$$
then
$$
(J+2I)^2= begin{bmatrix} 0 & 0 &&&\ &0&&&\ &&25 & * & * \ &&&25 &* \ &&&&25end{bmatrix} = mathrm {diag}((J(-2)+2I)^2, (J(3)+2I)^2),
$$
and $rank((J+2I)^2) = 3$, so
$$
rank((J(-2)+2I)^2) = 3 - rank((J(3)+2I)^2) = 3-3 = 0.
$$
answered Dec 5 '18 at 2:19
xbhxbh
5,7651522
What does $J(-2)$ mean? Also, since the dimension for diagonal entries $-2$ is $5$, how does this tell you that $rank(J(-2)^2)=1$? I know that this must involve the fact that $rank(C+2I)^2=6$ but I don't see how.
– Ryan Greyling
Dec 5 '18 at 4:42
@RyanGreyling $J(-2)$ is the biggest block with the diagonal entries $-2$. In this question, it is a upper triangular matrix with all diagonal entries $-2$. Example if the Jordan blocks are $J(3;4), J(3;5)$, then the $J(3) = mathrm {diag}(J(3;4), J(3;5))$.
– xbh
Dec 5 '18 at 4:47
@RyanGreyling For the 2nd question, note that for the Jordan form $J =mathrm {diag}(J(-2), J(3))$, $J+2I = mathrm {diag}(J(0), J(5))$, so the rank equals $mathrm {rank}(J(0)) + mathrm {rank}(J(5))$, where $J(5)$ is of full rank, i.e. 5.
– xbh
Dec 5 '18 at 4:51
Okay, I got it. The way I understand it uses the same logic in your explanation but with notation that makes more sense for me personally. Since $rank(C+2I)^2=6$, by rank-nullity theorem the dimension of the solution space for $(C+2I)^2vec{x}=0$ is $4$. $4<5$, the algebraic multiplicity for $lambda=-2$, so when drawing the box diagram for the Jordan form, there will be at least one column of $3$ boxes (one row above and one below). This means one of the Jordan blocks is of size $3$.
– Ryan Greyling
Dec 5 '18 at 19:03
add a comment |
Throughout the post, I use $J(c;m)$ to denote the Jordan block with diagonal entries $c$ of size $mtimes m$.
According to the char. polynomial, the Jordan form $J$ of $C$ has five $-2$ and five $3$ on the diagonal. For Jordan blocks with diagonal entries $3$, yes you have 2 blocks. For Jordan blocks with diag. entries $-2$, easy to see that $DeclareMathOperatorrank{rank} rank (J(-2;m)+2I) = m-1$ and $rank ((J(-2;m)+2I)^2) = m-2$. Restricting to the blocks with diagonal entries $-2$, the total dimension is $5$ by the char. polynomial, so $rank(J(-2)^2)=1$, then there exists at least $1$ block $J(-2;3)$, and the rest of the blocks could be 2 $J(-2;1)$ or 1 $J(-2;2)$.
Conclusion: up to permutation, the possible Jordan forms are $DeclareMathOperatordiag{diag} diag(J(-2;3), J(-2;2), J(3;1), J(3;4))$, $diag (J(-2;3), J(-2;2), J(3;2), J(3;3))$, $diag(J(-2;3), -2, -2, 3, J(3;4))$ and $diag (J(-2;3), -2, -2, J(3;2), J(3;3))$.
UPDATE
You could prove that
$$
rankbegin{bmatrix} A & O \O & Bend{bmatrix} = rank A + rank B
$$
where the LHS is a block diagonal matrix.
Then, for example,
$$
J = begin{bmatrix} -2 & 1 &&&\ &-2&&&\ &&3 & * & * \ &&&3 &* \ &&&&3end{bmatrix} = mathrm {diag}(J(-2), J(3)),
$$
then
$$
(J+2I)^2= begin{bmatrix} 0 & 0 &&&\ &0&&&\ &&25 & * & * \ &&&25 &* \ &&&&25end{bmatrix} = mathrm {diag}((J(-2)+2I)^2, (J(3)+2I)^2),
$$
and $rank((J+2I)^2) = 3$, so
$$
rank((J(-2)+2I)^2) = 3 - rank((J(3)+2I)^2) = 3-3 = 0.
$$
answered Dec 5 '18 at 2:19
xbhxbh
5,7651522
What does $J(-2)$ mean? Also, since the dimension for diagonal entries $-2$ is $5$, how does this tell you that $rank(J(-2)^2)=1$? I know that this must involve the fact that $rank(C+2I)^2=6$ but I don't see how.
– Ryan Greyling
Dec 5 '18 at 4:42
@RyanGreyling $J(-2)$ is the biggest block with the diagonal entries $-2$. In this question, it is a upper triangular matrix with all diagonal entries $-2$. Example if the Jordan blocks are $J(3;4), J(3;5)$, then the $J(3) = mathrm {diag}(J(3;4), J(3;5))$.
– xbh
Dec 5 '18 at 4:47
@RyanGreyling For the 2nd question, note that for the Jordan form $J =mathrm {diag}(J(-2), J(3))$, $J+2I = mathrm {diag}(J(0), J(5))$, so the rank equals $mathrm {rank}(J(0)) + mathrm {rank}(J(5))$, where $J(5)$ is of full rank, i.e. 5.
– xbh
Dec 5 '18 at 4:51
Okay, I got it. The way I understand it uses the same logic in your explanation but with notation that makes more sense for me personally. Since $rank(C+2I)^2=6$, by rank-nullity theorem the dimension of the solution space for $(C+2I)^2vec{x}=0$ is $4$. $4<5$, the algebraic multiplicity for $lambda=-2$, so when drawing the box diagram for the Jordan form, there will be at least one column of $3$ boxes (one row above and one below). This means one of the Jordan blocks is of size $3$.
– Ryan Greyling
Dec 5 '18 at 19:03
add a comment |
Throughout the post, I use $J(c;m)$ to denote the Jordan block with diagonal entries $c$ of size $mtimes m$.
According to the char. polynomial, the Jordan form $J$ of $C$ has five $-2$ and five $3$ on the diagonal. For Jordan blocks with diagonal entries $3$, yes you have 2 blocks. For Jordan blocks with diag. entries $-2$, easy to see that $DeclareMathOperatorrank{rank} rank (J(-2;m)+2I) = m-1$ and $rank ((J(-2;m)+2I)^2) = m-2$. Restricting to the blocks with diagonal entries $-2$, the total dimension is $5$ by the char. polynomial, so $rank(J(-2)^2)=1$, then there exists at least $1$ block $J(-2;3)$, and the rest of the blocks could be 2 $J(-2;1)$ or 1 $J(-2;2)$.
Conclusion: up to permutation, the possible Jordan forms are $DeclareMathOperatordiag{diag} diag(J(-2;3), J(-2;2), J(3;1), J(3;4))$, $diag (J(-2;3), J(-2;2), J(3;2), J(3;3))$, $diag(J(-2;3), -2, -2, 3, J(3;4))$ and $diag (J(-2;3), -2, -2, J(3;2), J(3;3))$.
UPDATE
You could prove that
$$
rankbegin{bmatrix} A & O \O & Bend{bmatrix} = rank A + rank B
$$
where the LHS is a block diagonal matrix.
Then, for example,
$$
J = begin{bmatrix} -2 & 1 &&&\ &-2&&&\ &&3 & * & * \ &&&3 &* \ &&&&3end{bmatrix} = mathrm {diag}(J(-2), J(3)),
$$
then
$$
(J+2I)^2= begin{bmatrix} 0 & 0 &&&\ &0&&&\ &&25 & * & * \ &&&25 &* \ &&&&25end{bmatrix} = mathrm {diag}((J(-2)+2I)^2, (J(3)+2I)^2),
$$
and $rank((J+2I)^2) = 3$, so
$$
rank((J(-2)+2I)^2) = 3 - rank((J(3)+2I)^2) = 3-3 = 0.
$$
answered Dec 5 '18 at 2:19
xbhxbh
5,7651522
Throughout the post, I use $J(c;m)$ to denote the Jordan block with diagonal entries $c$ of size $mtimes m$.
According to the char. polynomial, the Jordan form $J$ of $C$ has five $-2$ and five $3$ on the diagonal. For Jordan blocks with diagonal entries $3$, yes you have 2 blocks. For Jordan blocks with diag. entries $-2$, easy to see that $DeclareMathOperatorrank{rank} rank (J(-2;m)+2I) = m-1$ and $rank ((J(-2;m)+2I)^2) = m-2$. Restricting to the blocks with diagonal entries $-2$, the total dimension is $5$ by the char. polynomial, so $rank(J(-2)^2)=1$, then there exists at least $1$ block $J(-2;3)$, and the rest of the blocks could be 2 $J(-2;1)$ or 1 $J(-2;2)$.
Conclusion: up to permutation, the possible Jordan forms are $DeclareMathOperatordiag{diag} diag(J(-2;3), J(-2;2), J(3;1), J(3;4))$, $diag (J(-2;3), J(-2;2), J(3;2), J(3;3))$, $diag(J(-2;3), -2, -2, 3, J(3;4))$ and $diag (J(-2;3), -2, -2, J(3;2), J(3;3))$.
UPDATE
You could prove that
$$
rankbegin{bmatrix} A & O \O & Bend{bmatrix} = rank A + rank B
$$
where the LHS is a block diagonal matrix.
Then, for example,
$$
J = begin{bmatrix} -2 & 1 &&&\ &-2&&&\ &&3 & * & * \ &&&3 &* \ &&&&3end{bmatrix} = mathrm {diag}(J(-2), J(3)),
$$
then
$$
(J+2I)^2= begin{bmatrix} 0 & 0 &&&\ &0&&&\ &&25 & * & * \ &&&25 &* \ &&&&25end{bmatrix} = mathrm {diag}((J(-2)+2I)^2, (J(3)+2I)^2),
$$
and $rank((J+2I)^2) = 3$, so
$$
rank((J(-2)+2I)^2) = 3 - rank((J(3)+2I)^2) = 3-3 = 0.
$$
answered Dec 5 '18 at 2:19
xbhxbh
5,7651522
edited Dec 5 '18 at 5:17
answered Dec 5 '18 at 2:19
xbhxbh
5,7651522
answered Dec 5 '18 at 2:19
xbhxbh
5,7651522
answered Dec 5 '18 at 2:19
xbhxbh
5,7651522
5,7651522
What does $J(-2)$ mean? Also, since the dimension for diagonal entries $-2$ is $5$, how does this tell you that $rank(J(-2)^2)=1$? I know that this must involve the fact that $rank(C+2I)^2=6$ but I don't see how.
– Ryan Greyling
Dec 5 '18 at 4:42
@RyanGreyling $J(-2)$ is the biggest block with the diagonal entries $-2$. In this question, it is a upper triangular matrix with all diagonal entries $-2$. Example if the Jordan blocks are $J(3;4), J(3;5)$, then the $J(3) = mathrm {diag}(J(3;4), J(3;5))$.
– xbh
Dec 5 '18 at 4:47
@RyanGreyling For the 2nd question, note that for the Jordan form $J =mathrm {diag}(J(-2), J(3))$, $J+2I = mathrm {diag}(J(0), J(5))$, so the rank equals $mathrm {rank}(J(0)) + mathrm {rank}(J(5))$, where $J(5)$ is of full rank, i.e. 5.
– xbh
Dec 5 '18 at 4:51
Okay, I got it. The way I understand it uses the same logic in your explanation but with notation that makes more sense for me personally. Since $rank(C+2I)^2=6$, by rank-nullity theorem the dimension of the solution space for $(C+2I)^2vec{x}=0$ is $4$. $4<5$, the algebraic multiplicity for $lambda=-2$, so when drawing the box diagram for the Jordan form, there will be at least one column of $3$ boxes (one row above and one below). This means one of the Jordan blocks is of size $3$.
– Ryan Greyling
Dec 5 '18 at 19:03
add a comment |
What does $J(-2)$ mean? Also, since the dimension for diagonal entries $-2$ is $5$, how does this tell you that $rank(J(-2)^2)=1$? I know that this must involve the fact that $rank(C+2I)^2=6$ but I don't see how.
– Ryan Greyling
Dec 5 '18 at 4:42
@RyanGreyling $J(-2)$ is the biggest block with the diagonal entries $-2$. In this question, it is a upper triangular matrix with all diagonal entries $-2$. Example if the Jordan blocks are $J(3;4), J(3;5)$, then the $J(3) = mathrm {diag}(J(3;4), J(3;5))$.
– xbh
Dec 5 '18 at 4:47
@RyanGreyling For the 2nd question, note that for the Jordan form $J =mathrm {diag}(J(-2), J(3))$, $J+2I = mathrm {diag}(J(0), J(5))$, so the rank equals $mathrm {rank}(J(0)) + mathrm {rank}(J(5))$, where $J(5)$ is of full rank, i.e. 5.
– xbh
Dec 5 '18 at 4:51
Okay, I got it. The way I understand it uses the same logic in your explanation but with notation that makes more sense for me personally. Since $rank(C+2I)^2=6$, by rank-nullity theorem the dimension of the solution space for $(C+2I)^2vec{x}=0$ is $4$. $4<5$, the algebraic multiplicity for $lambda=-2$, so when drawing the box diagram for the Jordan form, there will be at least one column of $3$ boxes (one row above and one below). This means one of the Jordan blocks is of size $3$.
– Ryan Greyling
Dec 5 '18 at 19:03
What does $J(-2)$ mean? Also, since the dimension for diagonal entries $-2$ is $5$, how does this tell you that $rank(J(-2)^2)=1$? I know that this must involve the fact that $rank(C+2I)^2=6$ but I don't see how.
– Ryan Greyling
Dec 5 '18 at 4:42
What does $J(-2)$ mean? Also, since the dimension for diagonal entries $-2$ is $5$, how does this tell you that $rank(J(-2)^2)=1$? I know that this must involve the fact that $rank(C+2I)^2=6$ but I don't see how.
– Ryan Greyling
Dec 5 '18 at 4:42
@RyanGreyling $J(-2)$ is the biggest block with the diagonal entries $-2$. In this question, it is a upper triangular matrix with all diagonal entries $-2$. Example if the Jordan blocks are $J(3;4), J(3;5)$, then the $J(3) = mathrm {diag}(J(3;4), J(3;5))$.
– xbh
Dec 5 '18 at 4:47
@RyanGreyling $J(-2)$ is the biggest block with the diagonal entries $-2$. In this question, it is a upper triangular matrix with all diagonal entries $-2$. Example if the Jordan blocks are $J(3;4), J(3;5)$, then the $J(3) = mathrm {diag}(J(3;4), J(3;5))$.
– xbh
Dec 5 '18 at 4:47
@RyanGreyling For the 2nd question, note that for the Jordan form $J =mathrm {diag}(J(-2), J(3))$, $J+2I = mathrm {diag}(J(0), J(5))$, so the rank equals $mathrm {rank}(J(0)) + mathrm {rank}(J(5))$, where $J(5)$ is of full rank, i.e. 5.
– xbh
Dec 5 '18 at 4:51
@RyanGreyling For the 2nd question, note that for the Jordan form $J =mathrm {diag}(J(-2), J(3))$, $J+2I = mathrm {diag}(J(0), J(5))$, so the rank equals $mathrm {rank}(J(0)) + mathrm {rank}(J(5))$, where $J(5)$ is of full rank, i.e. 5.
– xbh
Dec 5 '18 at 4:51
Okay, I got it. The way I understand it uses the same logic in your explanation but with notation that makes more sense for me personally. Since $rank(C+2I)^2=6$, by rank-nullity theorem the dimension of the solution space for $(C+2I)^2vec{x}=0$ is $4$. $4<5$, the algebraic multiplicity for $lambda=-2$, so when drawing the box diagram for the Jordan form, there will be at least one column of $3$ boxes (one row above and one below). This means one of the Jordan blocks is of size $3$.
– Ryan Greyling
Dec 5 '18 at 19:03
Okay, I got it. The way I understand it uses the same logic in your explanation but with notation that makes more sense for me personally. Since $rank(C+2I)^2=6$, by rank-nullity theorem the dimension of the solution space for $(C+2I)^2vec{x}=0$ is $4$. $4<5$, the algebraic multiplicity for $lambda=-2$, so when drawing the box diagram for the Jordan form, there will be at least one column of $3$ boxes (one row above and one below). This means one of the Jordan blocks is of size $3$.
– Ryan Greyling
Dec 5 '18 at 19:03
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So you have that for $lambda =3$ we have 2 Jordan blocks. The question doesn't ask which ones exactly. Instead it asks which blocks they might be. Can you find the 2 possible configurations?
– I like Serena
Dec 4 '18 at 23:37
Suppose $C$ is already in Jordan normal form. Now pick some configuration of the blocks, say, 1 big block for $lambda=-2$. What will $(C+2I)^2$ look like? And what is its rank?
– I like Serena
Dec 4 '18 at 23:37
Hint: the number of Jordan blocks of an eigenvalue is equal to its geometric multiplicity.
– amd
Dec 5 '18 at 3:47