Frobenius norm of $n \times n$ matrix

Question

Find the Frobenius norm of the following $\mathbb{R}^{n\times n}$ matrix: \begin{bmatrix} 2 & 1 & 0 & 0\dots 0 \\ 1 & 2 & 1 & 0 \dots 0 \\ 0 & 1 & 2 & 1 \dots 0 \\ \vdots & &\ddots & \\ 0 & \ldots & 1 & 2 \end{bmatrix}

The Frobenius norm is the sum of squares of each element and then square root: I think it should be $\sqrt{n\cdot 2^2 + (n-1)\cdot 1 + (n-1) \cdot 1} $ since we have $n$ $2$s in the diagonal, and in the super and sub-diagonal we have $n-1$ $1$s. In total it is: $$\sqrt{4n+2n-2}= \sqrt{6n-2}$$

Is it correct?

Answer 1

Your matrix $A\in\mathbb{R}^{(n-1)\times n}$ has $n-1$ entries $(A)_{jj}=1$ for $j=1,...,n-1$ and $n-1$ entries $(A)_{j(j+1)}=2$ for $j=1,...,n-1$ and $n-2$ entries $(A)_{j(j+2)}=1$ for $j=1,...,n-2$. All other entries are zero and thus its Frobenius-norm is $$||A||_F=\sqrt{(n-1)1^2+(n-1)2^2+(n-2)1^2}=\sqrt{6n-7}.$$