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With the definition of matrix norm as

$$\|M \|=\sup_x \{ |Mx|: |x|=1 \},$$

where $M$ is square and $|\cdot|$ denotes the standard euclidean 2-norm. I'm trying to prove that

$$\|M\|^2=\|M^T\|^2 = \mathrm{largest \; eigenvalue \; of \;} M^TM?$$

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  • $\begingroup$ The title isn't accurate. $\endgroup$
    – Git Gud
    Mar 20 '14 at 15:51
  • $\begingroup$ @Git Gud - how's this? $\endgroup$ Mar 20 '14 at 16:39
  • $\begingroup$ The relevant thing in the question is proving it equals the largest eigenvalue, not that it equals the norm of the transpose (that will be an easy consequence). $\endgroup$
    – Git Gud
    Mar 20 '14 at 16:40
  • $\begingroup$ note that this is true only if M is orthogonally diagonalizable, for example M=[2 1 ; 0 1] $\endgroup$ Dec 21 '19 at 11:00
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Hint: note that $$ \|Mx\|^2 = (Mx)^T(Mx) = x^T(M^TM)x $$

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  • $\begingroup$ So similarly we have $$||M^Tx||^2 = x^TMM^T x,$$ I'm not sure where you would go from here though. $\endgroup$ Mar 20 '14 at 16:03
  • $\begingroup$ Re my previous comment, for real matrices $M$ we have $M^TM=MM^T$ ... $\endgroup$ Mar 20 '14 at 16:27
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    $\begingroup$ @DaveS: As Git Gud and copper.hat note, the next thing to note is that $M^TM$ and $MM^T$ always have the same eigenvalues, since $AB$ and $BA$ have the same eigenvalues for arbitrary matrices $A,B$. $\endgroup$ Mar 20 '14 at 23:06
  • $\begingroup$ I know that the derivative of $||x||_2^2$ w.r.t $x$ is equal to $2x$. But what about the derivative of $||x^T||_2^2$ w.r.t $x$. Is it the same $2x$? or $2x^T$? $\endgroup$
    – Christina
    Oct 27 '16 at 5:49
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    $\begingroup$ @DavidSimmons This is an old thread. But in case someone is confused by $M^TM=MM^T$ as I did. This is not true, and can be easily checked, e.g. Let $M$=[1 1; 0 0], where the ';' separates two rows. Then $M^TM$=[1 1; 1 1], but $MM^T$=[2 0; 0 0]. They do have the same eigenvalues though. $\endgroup$
    – syeh_106
    Nov 6 '19 at 9:32
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Show that the non zero eigenvalues of $AB$ and $BA$ are the same. (In this case $A=M,B=M^T$).

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