Questions about a linear mapping from $\mathbb{R^4}\rightarrow{}\mathbb{R^4}$

Question

My exam is in a couple hours and I wanted to clear a bit of confusion I have towards symmetric mappings, appreciate any help. Given is that $A:\mathbb{R^4}\rightarrow\mathbb{R^4}$ has, with respect to the standard basis of $\mathbb{R^4}$, the matrix,

$$ \left(\begin{matrix}1&-1&-1&-1\\-1&1&-1&-1\\-1&-1&1&-1\\-1&-1&-1&1\end{matrix}\right) $$

We are to find an orthonormal basis of eigenvectors of A.

Now am thinking because the matrix is symmetric, it has real eigenvalues and those eigenvectors with respect to different eigenvalues are orthogonal, so all I have to do here is find out the respective eigenvectors and apply gram-schmidt to the vectors corresponding to the same eigenvalue? What more can I say about this matrix? Or how can I tackle this problem best. Lastly, Saying that the mapping "with respect to the standard basis of $\mathbb{R^4}$", does that add any useable information to the question? Thanks a lot,

Answer 1

Your general approach is good, but in this case you don’t need to use Gram-Schmidt at all. An orthonormal eigenbasis can be found by inspection.

First, as Gerry Myerson noted in a comment, observe that all of the row sums are equal, so that $(1,1,1,1)^T$ is an eigenvector with eigenvalue $-2$. $A+2I$ has rank one, so the only other eigenvalue is $0$. So, you just have to find three pairwise orthogonal vectors that are orthogonal to $(1,1,1,1)^T$ and then normalize them all. Two fairly obvious choices are $(1,-1,0,0)^T$ and $(0,0,1,-1)^T$. If you can’t spot $(-1,-1,1,1)^T$ as a third orthogonal vector, you can at this stage use row-reduction to find a null vector of $$\begin{bmatrix}1&1&1&1\\1&-1&0&0\\0&0&1&-1\end{bmatrix},$$ which I think is still less work than applying Gram-Schmidt to complete the basis.

Knowing that the matrix is expressed relative to the standard basis is important for determining orthogonality. Implicit in the problem is that you’re using the standard Euclidean inner product, but its expression in terms of coordinates is basis-dependent. In particular, it only corresponds to the dot product of coordinate vectors when the basis is orthonormal, which the standard basis is.