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.



and let


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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