# Ngô Quốc Anh

## July 19, 2010

### On the determinant of a matrix

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 20:10

Several days ago, I placed a question on MathLinks asking the relation between $\det A$ and $\det(A-\lambda I)$. The point is how to evaluate $\displaystyle\det\begin{bmatrix}1+|x|^{2}-2x_{1}^{2}&-x_{1}x_{2}&\cdots&-x_{1}x_{n}\\ -x_{1}x_{2}&1+|x|^{2}-2x_{2}^{2}&\cdots&-x_{2}x_{n}\\ \vdots &\vdots &\ddots &\vdots\\ -x_{n}x_{1}&-x_{n}x_{2}&\cdots&1+|x|^{2}-2x_{n}^{2}\end{bmatrix}.$

Interestingly, K.M. showed me a new way to attack such a problem but slightly different from the original one. He proved $\displaystyle\det\begin{bmatrix}1+|x|^{2}-x_{1}^{2}&-x_{1}x_{2}&\cdots&-x_{1}x_{n}\\ -x_{1}x_{2}&1+|x|^{2}-x_{2}^{2}&\cdots&-x_{2}x_{n}\\ \vdots &\vdots &\ddots &\vdots\\ -x_{n}x_{1}&-x_{n}x_{2}&\cdots&1+|x|^{2}-x_{n}^{2}\end{bmatrix}=\left(1+|x|^{2}\right)^{n-1}.$

Let us discuss the proof of this modified problem.

Let $x=\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}$

and let $A=xx^T$.

The determinant we are trying to compute is $\displaystyle \det\left((1+|x|^2)I-A\right)$,

which is the characteristic polynomial of $A$ evaluated at $1+|x|^2$.

Now, $A$ is certainly diagonalizable (which doesn’t even matter, but it makes it easier to think about), and we know its eigenvalues. Why do we know its eigenvalues? Because $A$ is a matrix of rank 1, hence nullity $n-1$, hence $n-1$ of its $n$ eigenvalues are zero. What is the other eigenvalue? It’s the same as the sum of the eigenvalues, which is the trace of $A$, which is $|x|^2$. Put that information together, and we have that the characteristic polynomial of $A$ is $\det(\lambda I-A)=\left(\lambda-|x|^2\right)\lambda^{n-1}=\lambda^n-|x|^2\lambda^{n-1}$.

Substitute $1+|x|^2$ for $\lambda$ to get the result quoted.

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