Question about SVD proof from Trefethen and Bau
In Trefethen and Bau's proof of the SVD (see image below), they start by defining the following:
$$
U_1^* A V_1 = begin{bmatrix} sigma_1 & w^* \ 0 & B end{bmatrix}
$$
I understand the setup until this point, and I think I understand (broadly) how the proof proceeds: show that $w^* = 0$, then by the induction hypothesis, show that $B$ has the same structure, and so on.
What I don't understand is the block matrix above: why is the bottom-left element known to be zero, but the top right element isn't? Why don't we need to prove that the bottom left element is also 0?
linear-algebra svd
asked Nov 27 at 15:07
gwg
9201921
add a comment |
In Trefethen and Bau's proof of the SVD (see image below), they start by defining the following:
$$
U_1^* A V_1 = begin{bmatrix} sigma_1 & w^* \ 0 & B end{bmatrix}
$$
I understand the setup until this point, and I think I understand (broadly) how the proof proceeds: show that $w^* = 0$, then by the induction hypothesis, show that $B$ has the same structure, and so on.
What I don't understand is the block matrix above: why is the bottom-left element known to be zero, but the top right element isn't? Why don't we need to prove that the bottom left element is also 0?
linear-algebra svd
asked Nov 27 at 15:07
gwg
9201921
add a comment |
In Trefethen and Bau's proof of the SVD (see image below), they start by defining the following:
$$
U_1^* A V_1 = begin{bmatrix} sigma_1 & w^* \ 0 & B end{bmatrix}
$$
I understand the setup until this point, and I think I understand (broadly) how the proof proceeds: show that $w^* = 0$, then by the induction hypothesis, show that $B$ has the same structure, and so on.
What I don't understand is the block matrix above: why is the bottom-left element known to be zero, but the top right element isn't? Why don't we need to prove that the bottom left element is also 0?
linear-algebra svd
asked Nov 27 at 15:07
gwg
9201921
In Trefethen and Bau's proof of the SVD (see image below), they start by defining the following:
$$
U_1^* A V_1 = begin{bmatrix} sigma_1 & w^* \ 0 & B end{bmatrix}
$$
I understand the setup until this point, and I think I understand (broadly) how the proof proceeds: show that $w^* = 0$, then by the induction hypothesis, show that $B$ has the same structure, and so on.
What I don't understand is the block matrix above: why is the bottom-left element known to be zero, but the top right element isn't? Why don't we need to prove that the bottom left element is also 0?
linear-algebra svd
linear-algebra svd
asked Nov 27 at 15:07
gwg
9201921
asked Nov 27 at 15:07
gwg
9201921
asked Nov 27 at 15:07
gwg
9201921
asked Nov 27 at 15:07
gwg
9201921
asked Nov 27 at 15:07
gwg
9201921
9201921
add a comment |
add a comment |
1 Answer
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votes
We have constructed the bases so that $Av_1 = sigma_1 u_1$.
It follows that the first column of $U_1^*AV_1$ is
$$
(U_1^*AV_1)e_1 = U_1^*A(V_1e_1) = U_1^*Av_1 = U_1^* (sigma_1 u_1) = sigma_1 e_1
$$
where I have used $e_1$ to denote the first standard basis vector, $e_1 = (1,0,dots,0)^T$.
answered Nov 27 at 15:19
Omnomnomnom
126k788176
add a comment |
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1 Answer
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1 Answer
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We have constructed the bases so that $Av_1 = sigma_1 u_1$.
It follows that the first column of $U_1^*AV_1$ is
$$
(U_1^*AV_1)e_1 = U_1^*A(V_1e_1) = U_1^*Av_1 = U_1^* (sigma_1 u_1) = sigma_1 e_1
$$
where I have used $e_1$ to denote the first standard basis vector, $e_1 = (1,0,dots,0)^T$.
answered Nov 27 at 15:19
Omnomnomnom
126k788176
add a comment |
We have constructed the bases so that $Av_1 = sigma_1 u_1$.
It follows that the first column of $U_1^*AV_1$ is
$$
(U_1^*AV_1)e_1 = U_1^*A(V_1e_1) = U_1^*Av_1 = U_1^* (sigma_1 u_1) = sigma_1 e_1
$$
where I have used $e_1$ to denote the first standard basis vector, $e_1 = (1,0,dots,0)^T$.
answered Nov 27 at 15:19
Omnomnomnom
126k788176
add a comment |
We have constructed the bases so that $Av_1 = sigma_1 u_1$.
It follows that the first column of $U_1^*AV_1$ is
$$
(U_1^*AV_1)e_1 = U_1^*A(V_1e_1) = U_1^*Av_1 = U_1^* (sigma_1 u_1) = sigma_1 e_1
$$
where I have used $e_1$ to denote the first standard basis vector, $e_1 = (1,0,dots,0)^T$.
answered Nov 27 at 15:19
Omnomnomnom
126k788176
We have constructed the bases so that $Av_1 = sigma_1 u_1$.
It follows that the first column of $U_1^*AV_1$ is
$$
(U_1^*AV_1)e_1 = U_1^*A(V_1e_1) = U_1^*Av_1 = U_1^* (sigma_1 u_1) = sigma_1 e_1
$$
where I have used $e_1$ to denote the first standard basis vector, $e_1 = (1,0,dots,0)^T$.
answered Nov 27 at 15:19
Omnomnomnom
126k788176
edited Nov 27 at 15:25
answered Nov 27 at 15:19
Omnomnomnom
126k788176
answered Nov 27 at 15:19
Omnomnomnom
126k788176
answered Nov 27 at 15:19
Omnomnomnom
126k788176
126k788176
add a comment |
add a comment |
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