# Squared norm of matrix equal to squared norm of its transpose

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 title isn't accurate. Mar 20 '14 at 15:51
• @Git Gud - how's this? Mar 20 '14 at 16:39
• 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). Mar 20 '14 at 16:40
• note that this is true only if M is orthogonally diagonalizable, for example M=[2 1 ; 0 1] Dec 21 '19 at 11:00

Hint: note that $$\|Mx\|^2 = (Mx)^T(Mx) = x^T(M^TM)x$$
• So similarly we have $$||M^Tx||^2 = x^TMM^T x,$$ I'm not sure where you would go from here though. Mar 20 '14 at 16:03
• Re my previous comment, for real matrices $M$ we have $M^TM=MM^T$ ... Mar 20 '14 at 16:27
• @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 $A,B$. 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 $2x^T$? Oct 27 '16 at 5:49
• @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. Nov 6 '19 at 9:32
Show that the non zero eigenvalues of $AB$ and $BA$ are the same. (In this case $A=M,B=M^T$).