Matrix/vector : (A-B)s = 0 does not imply A=B

  • Thread starter elgen
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  • #1
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Main Question or Discussion Point

Let s be a column vector and L be a diagonal matrix. Then

[tex]s^T L s L s = L s s^T L s [/tex]

by noticing that [tex] s^TLs[/tex] is a scalar.

However,

[tex] s^T L s L \neq L s s^T L [/tex].

Is there some mathematical explanation behind this? Thank you.

elgen
 

Answers and Replies

  • #2
Fredrik
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Are you looking for a reason why Ax=0 doesn't imply x=0, when A is a non-zero linear operator and x a vector (or equivalently, when A is a m×n matrix and x an n×1 matrix)? One way of looking at it is that there are non-zero linear operators that aren't invertible. This is different from how real numbers behave, since every non-zero real number r has the multiplicative inverse 1/r. If A is invertible, we clearly have x=0. But we can easily find examples of non-invertible, non-zero A such that Ax=0 with x≠0.

[tex]\begin{pmatrix}0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 \\ 0\\ 0\end{pmatrix}=0[/tex]

Here A projects onto the 23 plane, and x is in the 1 direction.
 
  • #3
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Thank you for replying my message, Fredrik. Even though it is not the expected answer, it does explain well the presence of the NULL space of the matrix A.

What has surprised me is that, let [tex]\alpha=s^TL s [/tex], the equality is really
[tex]\alpha Ls = L s \alpha [/tex].

Since
[tex]s^T L s Ls = L s s^T L s[/tex]
we have
[tex](s^T Ls L - L s s^T L)s = 0 [/tex]

This does not imply that [tex] s^T L s L = L s s^T L[/tex].
 
  • #4
Fredrik
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Right. I answered the question in the title (or rather, one that's easily seen to be completely equivalent to it), and the answer should explain the issue you brought up in the post (the one you're repeating now). If sTLsL-LssTL had been invertible, we would obviously have had s=0. The fact s≠0 is a possibility means that we have found an example (sTLsL-LssTL) of a non-zero, non-invertible linear operator.
 
  • #5
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Thank you for the discussion.
 

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