Prove that exists a subspace $U$ of $M_{4\times4}^{\mathbb{R}}$ such that the length of the orthogonal projection of matrix $A$ onto $U$ is 5

Question

Prove that exists a subspace $U$ of $M_{4\times4}^{\mathbb{R}}$ such that the length of the orthogonal projection of matrix $A$ onto $U$ is 5.

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

And the inner product is defined as:

$<A,B>=Trace(AB^t)$

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I did some calculations and I've found that the subspace

$U = Sp\{\begin{bmatrix} 1 + \sqrt{10} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$}

matches the requirements, is that correct? if not, can someone explain how does someone approaches this problem generally?

Answer 1

My approach would start with the calculation of $\|A\|$, where $\|\cdot\|$ as defined above is the Frobenius norm (often called Hilbert-Schmidt norm in the infinite-dimensional situation). Then recall the general fact that every orthogonal projection $P\neq 0$ satisfies $\|P\|=1$.

In the given situation $A$ were thus required to have norm at least equal to $5$, if such a subspace $U$ indeed exists.

As you are quite parsimonious to present your calculations$-$It is generally considered helpful to include them$-$I'm refraining from presenting more details ...