Linear algebra by Loehr exercise

Question

Given $a,b\in R$ evaluate $\det(aI_n+bJ_n)$, where $J_n\in M_n(R)$ has every entry equal to $1_R$.

I'm studying algebra, the book I use leaves this exercise, but I'm not able to solve it, with induction. $\mathbb{1}_{R}: R \to R$ defined by $\mathbb{1}_{R}(x) = b$
I have that doubt, who can answer it please.

Hint: What’s the relationship between the determinant of a matrix and its eigenvalues? — amd, Feb 14 '18 at 01:38
Please take the time to enter critical parts of your question as text instead of an image. The latter isn’t searchable, doesn’t show up in summaries, nor is it accessible to people using screen readers. Typing in one more sentence couldn’t have taken more time and effort than taking a photo/screen shot, cropping it, uploading it and linking it to your question. — amd, Feb 14 '18 at 01:41
Particular case of https://math.stackexchange.com/a/2025336/ . — darij grinberg, Feb 14 '18 at 02:19
See also: Determinant of a specially structured matrix ($a$'s on the diagonal, all other entries equal to $b$) — Martin Sleziak, Apr 21 '19 at 08:39

score 2 · Accepted Answer · answered Feb 14 '18 at 02:11

Start by finding the eigenvalues and multiplicities of $J_n$. If you compute eigenvalues for a few of these matrices (Wolfram Alpha might help you here), it should become clear that $0$ is an eigenvalue with multiplicity $n - 1$ and $n$ is an eigenvalue with multiplicity $1$.

To show $J_n$ has $0$ as an eigenvalue with multiplicity $n - 1$, I suggest computing the rank of $J_n$. Since every column (or row) vector is the same, the span of the columns is of dimension $1$. By the rank-nullity theorem, the nullspace of $J_n$, which is the dimension of the eigenspace corresponding to $0$, is going to be $n - 1$.

For the eigenvalue of $n$, I suggest computing the eigenvector corresponding to this eigenvalue. This should reveal that the eigenvector ought to be a multiple of $(1, 1, \ldots, 1)^T$. And, indeed, multiplying this vector to $J_n$ clearly produces $n$ times this vector, proving $n$ is indeed an eigenvalue for $J_n$.

Since we have found multiplicities of eigenvalues totaling $n$, there can be no others.

Now, $bJ_n$ has eigenvalues with the same multiplicities, but now multiplied by $b$. If we add $aI_n$ to a matrix, this adds $a$ to each eigenvalue. So, the eigenvalues of $aI_n + bJ_n$ are $a$ with multiplicity $n - 1$ and $a + bn$ with multiplicity $1$. This produces a determinant $$a^{n-1}(a + bn).$$

Linear algebra by Loehr exercise

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