# Orthogonal matrices and matrix norms

I have seen some disagreement online and was wondering if anyone could clarify for me:

If $$X$$ is an arbitrary $$n \times n$$ matrix and $$A$$ is an arbitrary orthogonal $$n \times n$$ matrix, is it true that $$\| AX \|_p = \|X\|_p$$

For all $$p \in \mathbb{Z}_+\cup{\infty}$$, where $$\|\cdot\|_p$$ is the matrix $$p$$-norm defined as:

$$\| A \|_p = \sup_{x \neq 0} \frac{|Ax|_p}{|x|_p}$$

Where $$|\cdot|_p$$ is the $$p$$-norm (of vector).

• Ah damN! Sorry! – JDoe2 Jun 3 at 13:07
• $\|\cdot\|_p$ isn't a norm when $p<1$. – user1551 Jun 3 at 13:08
• Apologies yes that has been rectified also. – JDoe2 Jun 3 at 13:11

Proposition: $$\left\lVert UAW\right\lVert=||A||$$ if $$U$$ and $$W$$ are orthogonal or unitary, for the Frobenius norm and for the operator norm induced by the vector norm $$\left\lVert\cdot\right\lVert_2.$$

\begin{align*} \left\lVert UAW\right\lVert_F&=\text{tr} (UAWW^TA^TQ^T)=\text{tr}(UAA^TU^T)\\ &=\text{tr}(A^TU^TU^TA)=\text{tr}(AA^T)\\ &=\left\lVert A\right\lVert_F\end{align*}.

Next, for the operator norm induced by the $$2$$-norm:

\begin{align*} \left\lVert UAW\right\lVert_2 &=\max_{x\neq 0}\frac{\left\lVert UAWx \right\lVert_2}{\left\lVert x\right\lVert_2}=\max_{x\neq 0}\frac{\sqrt{x^TW^TA^TU^TUAWx}}{\sqrt{x^Tx}}\\ &=\max_{z\neq 0}\frac{\sqrt{zA^TAz}}{\sqrt{z^Tz}}=\max_{z\neq 0}\frac{\left\lVert Az\right\lVert_2}{\left\lVert z\right\lVert_2}=\left\lVert A\right\lVert_2, \end{align*} where we used the substitution $$z=Wx.$$

For an example to show that the infinity and $$1$$-norm do not work, take $$U$$ to be rotation matrix which rotates counterclockwise by $$60$$ degrees, $$A=\begin{bmatrix} 1 & 2 \\ 3 & 4\end{bmatrix},$$ and $$W=I$$. Then, $$UAW=\begin{bmatrix} \frac{1}{2}-\frac{3\sqrt{3}}{2} & 1-2\sqrt{3} \\ \frac{3}{2}+\frac{\sqrt{3}}{2} & 2+\sqrt{3}\end{bmatrix}.$$

• Can I confirm, are you saying these are the only two matrix norms for which this holds? For example does this hold for the 1-norm and Infinite-norm? – JDoe2 Jun 3 at 13:10
• Yes, we want our norm induced by an inner product (and $\ell_2$ is the only Hilbert space for all $\ell_p$). The Frobenius norm also comes from an inner product, namely $(A,B)=\text{tr}(B^*A).$ – cmk Jun 3 at 13:16
• Also, if you want an example, just take $A=\begin{bmatrix} 1 & 2\\ 3 & 4\end{bmatrix}$, $W=I$, and $U$ to be a rotation by, say, $\pi/3$. – cmk Jun 3 at 13:22

Why would that be true for any matrices $$A,X$$ in general. It is true that

$$\| AX \|_{p} \leq ||A\|_{p} \|X\|_{p}$$

when $$p=2$$ and $$A$$ is orthogonal then this is true

$$|| A X ||_{2} = ||X||_{2}$$

the $$2$$ norm is invariant for orthogonal matrices. If you're looking for a proof I'm pretty sure this has been asked before.

Here is a counter example. Suppose that $$A=X$$ and

$$A = \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}$$

and let $$\theta = \frac{\pi}{4}$$ $$A = \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix}$$

$$A \cdot A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$$

and the $$1$$ norm is given as

$$\|A\|_{1} = \max_{1 \leq j \leq n} \sum_{i=1}^{m} |a_{ij}|$$

is the max column sum of $$A$$

$$\|A\|_{1} = \frac{2}{\sqrt{2}}$$

$$\|B\|_{1} = 1$$

• Apologies I missed the word orthogonal in the question, it has now been rectified. – JDoe2 Jun 3 at 13:09
• I am looking for more general than that, not just 2 but any p. – JDoe2 Jun 3 at 13:10
• Its only true for $p=2$ – Shogun Jun 3 at 13:11
• Perhaps could you give a counterexample for $p=1$ or $p=\infty$? – JDoe2 Jun 3 at 13:14
• In your example, $A$ is not orthogonal. – cmk Jun 3 at 13:36