Squared norm of matrix equal to squared norm of its transpose

Question

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

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

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). — Git Gud, Mar 20 '14 at 16:40
note that this is true only if M is orthogonally diagonalizable, for example M=[2 1 ; 0 1] — Omer Ben Haim, Dec 21 '19 at 11:00

score 13 · Accepted Answer · answered Mar 20 '14 at 15:50

13

Hint: note that $$ \|Mx\|^2 = (Mx)^T(Mx) = x^T(M^TM)x $$

answered Mar 20 '14 at 15:50

Ben Grossmann

So similarly we have
$$||M^Tx||^2 = x^TMM^T x,$$ I'm not sure where you would go from here though.
– David Simmons Mar 20 '14 at 16:03
Re my previous comment, for real matrices $M$ we have $M^TM=MM^T$ ... – David Simmons Mar 20 '14 at 16:27
2

@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 $A,B$. – Ben Grossmann Mar 20 '14 at 23:06
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 $2x^T$? – Christina Oct 27 '16 at 05:49
5

@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. – syeh_106 Nov 06 '19 at 09:32

score 6 · Answer 2 · answered Mar 20 '14 at 15:33

6

Show that the non zero eigenvalues of $AB$ and $BA$ are the same. (In this case $A=M,B=M^T$).

answered Mar 20 '14 at 15:33

copper.hat

2 Answers2