Lemma relating linear independence to subspaces. How? (Suppose x…xs are vectors which span W and y…yt are...
$begingroup$
Sorry for the absolutely horrid title, didn't want to have a vague one. Here's the full lemma:
Let W$subset R^n$ be a subspace Suppose that ${u_1,u_2}$...${u_s}$ are vectors which span W, and ${v_1,v_2}$...${v_t}$ are vectors in W which are linearly independent. Then s $ge$ t.
Why is the above lemma true? Furthermore, how does it lead us to the definition of the dimension of W? The dimension of W is the number of vectors that make up a basis of W, but the vectors v in the lemma don't necessarily make up the basis do they (it's not clarified whether they span W or not)?
But from the looks of the question, we are meant to define the dimension using this lemma. How?
linear-algebra independence
asked Dec 7 '18 at 23:23
James RonaldJames Ronald
1007
$endgroup$
add a comment |
$begingroup$
Sorry for the absolutely horrid title, didn't want to have a vague one. Here's the full lemma:
Let W$subset R^n$ be a subspace Suppose that ${u_1,u_2}$...${u_s}$ are vectors which span W, and ${v_1,v_2}$...${v_t}$ are vectors in W which are linearly independent. Then s $ge$ t.
Why is the above lemma true? Furthermore, how does it lead us to the definition of the dimension of W? The dimension of W is the number of vectors that make up a basis of W, but the vectors v in the lemma don't necessarily make up the basis do they (it's not clarified whether they span W or not)?
But from the looks of the question, we are meant to define the dimension using this lemma. How?
linear-algebra independence
asked Dec 7 '18 at 23:23
James RonaldJames Ronald
1007
$endgroup$
add a comment |
$begingroup$
Sorry for the absolutely horrid title, didn't want to have a vague one. Here's the full lemma:
Let W$subset R^n$ be a subspace Suppose that ${u_1,u_2}$...${u_s}$ are vectors which span W, and ${v_1,v_2}$...${v_t}$ are vectors in W which are linearly independent. Then s $ge$ t.
Why is the above lemma true? Furthermore, how does it lead us to the definition of the dimension of W? The dimension of W is the number of vectors that make up a basis of W, but the vectors v in the lemma don't necessarily make up the basis do they (it's not clarified whether they span W or not)?
But from the looks of the question, we are meant to define the dimension using this lemma. How?
linear-algebra independence
asked Dec 7 '18 at 23:23
James RonaldJames Ronald
1007
$endgroup$
Sorry for the absolutely horrid title, didn't want to have a vague one. Here's the full lemma:
Let W$subset R^n$ be a subspace Suppose that ${u_1,u_2}$...${u_s}$ are vectors which span W, and ${v_1,v_2}$...${v_t}$ are vectors in W which are linearly independent. Then s $ge$ t.
Why is the above lemma true? Furthermore, how does it lead us to the definition of the dimension of W? The dimension of W is the number of vectors that make up a basis of W, but the vectors v in the lemma don't necessarily make up the basis do they (it's not clarified whether they span W or not)?
But from the looks of the question, we are meant to define the dimension using this lemma. How?
linear-algebra independence
linear-algebra independence
asked Dec 7 '18 at 23:23
James RonaldJames Ronald
1007
asked Dec 7 '18 at 23:23
James RonaldJames Ronald
1007
edited Dec 7 '18 at 23:35
James Ronald
asked Dec 7 '18 at 23:23
James RonaldJames Ronald
1007
asked Dec 7 '18 at 23:23
James RonaldJames Ronald
1007
asked Dec 7 '18 at 23:23
James RonaldJames Ronald
1007
1007
add a comment |
add a comment |
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