Eigen do this one!

Algebra Level 2

Let $A$ be the matrix $A = \begin{pmatrix} 3 & 5 \\ 1 & 7 \\ \end{pmatrix}$ . An eigenvalue of $A$ is a real number $\lambda$ such that there exists a non-zero vector ${x_\lambda} = \begin{pmatrix} X \\ Y \end{pmatrix}$ such that $Ax_\lambda=\lambda x_\lambda$ . What is the sum of all the eigenvalues of $A$ (i.e. sum of all the possible $\lambda$ )?

The answer is 10.

1 solution

Arron Kau Staff
May 13, 2014

Solution 1: $Ax = \lambda x \Rightarrow (A-\lambda I)x = 0$ , where $I$ is the identity matrix. Since $x$ is not the zero vector, this implies that the determinant of $A-\lambda I$ is 0. So $\begin{aligned} 0=|A - \lambda I| &= \begin{vmatrix} 3 - \lambda & 0 \\ 1 & 10 - \lambda \\ \end{vmatrix} \\ &= (3 - \lambda)(10 - \lambda) - (0)(1) \\ &= \lambda^2 - 10 \lambda -3\lambda + 30 - (0) \\ &= \lambda^2 - 13 \lambda + 30 \\ &= (\lambda - 10)(\lambda - 3) \\ \end{aligned}$

Thus the eigenvalues of $A$ are $10$ and $3$ . Hence the sum is $10 + 3 = 13$ .

Solution 2: From Linear Algebra, the sum of all the (complex) eigenvalues is equal to the trace of the matrix (sum of entires on the main diagonal). This fact follows from applying Vieta's formula to the polynomial $\det(A-\lambda I)$ . Hence, the sum of all the eigenvalues is $3+10=13$ .

Note: The existence of vectors $x_10$ and $x_3$ are guaranteed from theory. They may be obtained by setting $X=1$ .

Looks like the solution given here (13) is different from the answer (10)

Brian Blass - 1 year, 5 months ago

Eigen do this one!

The answer is 10.

1 solution

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