# Is the spectral norm of a matrix greater than the spectral norm of a top principle submatrix of it?

I have the following question, which I need for a research work:

Suppose that $$A$$ is an $$n \times n$$ symmetric, positive definite matrix, and let $$m$$ be a positive integer less than $$n$$. Let $$B$$ be the $$m \times m$$ top principle submatrix of $$A$$, i.e. $$B := (A_{ij})_{i\le m, j\le m}$$. Then, is the spectral norm (largest magnitude eigenvalue) of $$B$$ less than the spectral norm of $$A$$? If not true, can a counterexample be given?

• Are you sure you are asking what you intended to ask? The so-called "spectral norm" of a matrix is not the largest magnitude of all eigenvalues (which is known as the spectral radius), but the largest singular value of that matrix. Commented May 7, 2021 at 7:11
• For the record, you are talking about properties of $A$ that are base-invariant, so $B$ being that specific principal minor seems irrelevant.
– user239203
Commented May 7, 2021 at 7:11
• @user1551, I am talking about the spectral norm, which is the largest singular value, but in my case, the matrix is symmetric (I forgot to mention that), so the spectral norm is indeed the largest magnitude eigenvalue. Commented May 7, 2021 at 7:19

## 1 Answer

This is true for any complex square matrix, not just for the real symmetric ones. It follows directly from the characterisation of the spectral norm that $$\|A\|_2=\max_{\|x\|_2=1}\|Ax\|_2$$. More specifically, let $$y\in\mathbb C^m$$ be a unit vector such that $$\|By\|_2=\|B\|_2$$. Append zeroes to $$y$$ to form a unit vector $$x\in\mathbb C^n$$. Then $$\|B\|_2=\|By\|_2\le\|Ax\|_2\le\|A\|_2$$.

As a remark, note that unless the matrix is Hermitian, the spectral radius of a principal submatrix can be larger than the spectral radius matrix of the parent matrix. E.g. in the companion matrix $$A=\pmatrix{0&-1\\ 1&2},$$ the spectral radius of the trailing principal submatrix $$2$$ is greater than $$1$$, the spectral radius of $$A$$.