Frobenius norm and operator norm inequality

Question

Let $A$ be a $k\times m$ matrix and B be a $m\times n$ matrix, I wonder how to prove the following inequality

$$\|AB\|_F\le\|A\| \|B\|_F,$$

where $\|\cdot\|_F$ is the Frobenius norm (square root of the sum of all squared entries and $\|\cdot\|$ is the 2-operator norm )

Note if $n=1$, i.e when $B$ is a column vector, this just follows from the definition of the operator norm. But I don't know how to deal with the general case. I have thought about using SVD of $A,B$ but don't know how to simplify the LHS. Any approach will be appreciated!

user8675309 · Accepted Answer · 2020-02-08T05:35:03.470

6

if you know a little spectral theory, you can square both sides and recognize your problem is equivalent to proving

$\text{trace}\big(XY\big) \leq \lambda_1 \cdot \text{trace}\big(Y\big)$
here $\lambda_1$ is the maximal eigenvalue of $X$ and $X, Y$ are Hermitian positive semidefinite. $X$ is unitarily diagonalizable by $Q$ so

$\text{trace}\Big(XY\Big) $
$=\text{trace}\Big(Q\Lambda Q^*Y\Big) $
$=\text{trace}\Big(\Lambda \big(Q^*YQ\big)\Big) $
$=\text{trace}\Big(\Lambda Z\Big) $
$=\sum_{k} \lambda_k \cdot z_{k,k}$
$\leq \sum_{k} \lambda_1 \cdot z_{k,k}$
$= \lambda_1 \cdot \text{trace}\Big(Z\Big)$
$= \lambda_1 \cdot \text{trace}\Big(Q^* Y Q\Big)$
$= \lambda_1 \cdot \text{trace}\Big(Y\Big)$

selecting $X:= A^*A$ and $Y:= BB^*$ completes the proof

edited Feb 08 '20 at 05:35

answered Feb 08 '20 at 04:49

user8675309

12,193

May I have a question at equation 4? It shoud be the sum of singular value of $X$ times $z_{k,k}$ instead of eigenvalue. – Tung Nguyen Sep 09 '21 at 17:07
1

@TungNguyen you should try writing down the relationship between singular values and eigenvalues for Hermitian positive semi-definite matrices. – user8675309 Sep 09 '21 at 18:49
Could you explain why $tr((AB)^AB)=tr(A^ABB^*)$ and also the second equality? – Michael Novak May 02 '25 at 06:48
1

@MichaelNovak I think you are looking for cyclic property of trace – user8675309 May 02 '25 at 15:12
@user8675309 Thank you very much! I never heard of this property. Is it something one typically learn in a linear algebra course? – Michael Novak May 02 '25 at 15:28

Conifold · Answer 2 · 2020-02-09T00:57:19.887

Think of the matrix $B$ as a long vector listing its columns in sequence, so it is an $n^2$-dimensional vector assuming $A,B$ are $n\times n$. Then $\|B\|_F=\text{tr}(B^*B)=\sum_{i,j}|b_{ij}|^2$ is just the Euclidean norm of this long vector. Now, when we multiply $B$ by $A$ the resulting matrix has $Ab^{(j)}$ as columns, where $b^{(j)}$ are the columns of $B$, and $\|Ab^{(j)}\|\leq\|A\|\|b^{(j)}\|$ for each $j$, where $\|A\|$ is the spectral norm.

But this means that $A$ induces a linear map on our long vectors for $B$, with each column segment of $n$ coordinates multiplied by $A$. The norm of this map $\leq\|A\|$ (actually, it is equal to it, but we do not need that), because the above inequality holds for each column segment. But then $\|AB\|_F\le\|A\| \|B\|_F$ is just the usual inequality $\|Ax\|\leq\|A\|\|x\|$ for the Euclidean norms of vectors, only applied to the long vectors.

Frobenius norm and operator norm inequality

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