On the Determinant and the Trace of a matrix

Suppose you have a squared matrix A with n rows and columns. Is there a relationship between its eigenvalues and its determinant? And what about its trace?

Determinant of A.

Consider the determinant of \mid A- \lambda I \mid, which is a polynomial of degree n. The values of \lambda that solve the equation \mid A- \lambda I \mid = 0, are the eigenvalues of A, and are n as its degree.
We can write such a polynomial using its roots (its eigenvalues) as follows:
\mid A- \lambda I \mid = (\lambda - \lambda_1)(\lambda - \lambda_2)\cdots(\lambda - \lambda_n)

Its constant term is: (-1)^n \lambda_1\cdots\lambda_n,
but it’s also the value of \mid A- \lambda I \mid with \lambda = 0, which obviously is \mid A \mid.

So \mid A \mid = (-1)^n \lambda_1\cdots\lambda_n.

Trace of A.

If we develop the term of degree n-1 we obtain:
-(\lambda_1+\lambda_2+\cdots\+lambda_n).

For simplicity we consider the case of an 3x3 matrix:

\mid A- \lambda I \mid = \begin{bmatrix}a_{11}-\lambda & a_{12} & a_{13} \\a_{21} & a_{22}-\lambda &  a_{23} \\a_{31} & a_{32} & a_{33}-\lambda \end{bmatrix}

It can be decomposed as follows:

\mid A- \lambda I \mid = \begin{bmatrix}a_{11} & a_{12} & a_{13} \\a_{21} & a_{22}-\lambda & a_{23} \\a_{31} & a_{32} & a_{33}-\lambda \end{bmatrix}+\begin{bmatrix}-\lambda & 0 & 0 \\a_{21} & a_{22}-\lambda & a_{23} \\a_{31} & a_{32} & a_{33}-\lambda \end{bmatrix}

Now we can apply the decomposition on the second row of the second matrix:

\begin{bmatrix}-\lambda & 0 & 0 \\a_{21} & a_{22}-\lambda & a_{23} \\a_{31} & a_{32} & a_{33}-\lambda \end{bmatrix}=\begin{bmatrix}-\lambda & 0 & 0 \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33}-\lambda \end{bmatrix}+\begin{bmatrix}-\lambda & 0 & 0 \\0 & -\lambda & 0 \\a_{31} & a_{32} & a_{33}-\lambda \end{bmatrix}

And finally we can apply the decomposition on the third row of the last matrix:


\begin{bmatrix}-\lambda & 0 & 0 \\0 & -\lambda & 0 \\a_{31} & a_{32} & a_{33}-\lambda \end{bmatrix}=\begin{bmatrix}-\lambda & 0 & 0 \\0 & -\lambda & 0 \\a_{31} & a_{32} & a_{33} \end{bmatrix}+\begin{bmatrix}-\lambda & 0 & 0 \\0 & -\lambda & 0 \\0 & 0 & -\lambda \end{bmatrix}

The matrices that give a contribute to the n-1 degree of \mid A- \lambda I \mid are:

D1=\begin{bmatrix}a_{11} & a_{12} & a_{13} \\a_{21} & a_{22}-\lambda & a_{23} \\a_{31} & a_{32} & a_{33}-\lambda \end{bmatrix}

D2=\begin{bmatrix}-\lambda & 0 & 0 \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33}-\lambda \end{bmatrix}

D3=\begin{bmatrix}-\lambda & 0 & 0 \\0 & -\lambda & 0 \\a_{31} & a_{32} & a_{33} \end{bmatrix}

So their n-1 degree contribute is: (a_{11}+a_{22}+a_{33}).

Generalizing the contribution is: (-1)^{n-1}(a_{11}+\cdots+a_{nn}).

Finally Trace(A)=\sum_1^n a_{ii}=(-1)^n\sum_1^n \lambda_{j}

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