GRE Quantitative Reasoning - Eigenspace Classifications

Find the eigenvalues of the following matrix, if possible.

$A=\begin{bmatrix} 1&2 \ -2&1 \end{bmatrix}$

$\lambda_{1,2}=1\pm2i$

$\lambda_{1,2}=\pm1$

$\lambda=\frac{5}{2}$

$\lambda=\begin{bmatrix} 2i&4i \ -2i&i \end{bmatrix}$

The eigenvalues do not exist.

Explanation

In order to find the eigenvalues of a matrix, apply the following formula:

$det(A-\lambda I) =0$

is the identity matrix.

$\begin{bmatrix} 1&2 \ -2&1 \end{bmatrix}-\lambda \begin{bmatrix} 1&0 \ 0&1 \end{bmatrix}=\begin{bmatrix} 1-\lambda & 2\ -2&1-\lambda \end{bmatrix}$

Compute the determinant and set it equal to zero.

$(1-\lambda)(1-\lambda)-(2)(-2)=0$

$1-2\lambda+\lambda^2+4=0$

$\lambda^2-2\lambda+5=0$

Solve for lambda by using the quadratic formula.

$\lambda_{1,2}=1\pm2i$

Find the eigenvalues of the following matrix, if possible.

$A=\begin{bmatrix} 1&2 \ -2&1 \end{bmatrix}$

$\lambda_{1,2}=1\pm2i$

$\lambda_{1,2}=\pm1$

$\lambda=\frac{5}{2}$

$\lambda=\begin{bmatrix} 2i&4i \ -2i&i \end{bmatrix}$

The eigenvalues do not exist.

Explanation

In order to find the eigenvalues of a matrix, apply the following formula:

$det(A-\lambda I) =0$

is the identity matrix.

$\begin{bmatrix} 1&2 \ -2&1 \end{bmatrix}-\lambda \begin{bmatrix} 1&0 \ 0&1 \end{bmatrix}=\begin{bmatrix} 1-\lambda & 2\ -2&1-\lambda \end{bmatrix}$

Compute the determinant and set it equal to zero.

$(1-\lambda)(1-\lambda)-(2)(-2)=0$

$1-2\lambda+\lambda^2+4=0$

$\lambda^2-2\lambda+5=0$

Solve for lambda by using the quadratic formula.

$\lambda_{1,2}=1\pm2i$

Find the eigenvalues of the following matrix, if possible.

$A=\begin{bmatrix} 1&2 \ -2&1 \end{bmatrix}$

$\lambda_{1,2}=1\pm2i$

$\lambda_{1,2}=\pm1$

$\lambda=\frac{5}{2}$

$\lambda=\begin{bmatrix} 2i&4i \ -2i&i \end{bmatrix}$

The eigenvalues do not exist.

Explanation

In order to find the eigenvalues of a matrix, apply the following formula:

$det(A-\lambda I) =0$

is the identity matrix.

$\begin{bmatrix} 1&2 \ -2&1 \end{bmatrix}-\lambda \begin{bmatrix} 1&0 \ 0&1 \end{bmatrix}=\begin{bmatrix} 1-\lambda & 2\ -2&1-\lambda \end{bmatrix}$

Compute the determinant and set it equal to zero.

$(1-\lambda)(1-\lambda)-(2)(-2)=0$

$1-2\lambda+\lambda^2+4=0$

$\lambda^2-2\lambda+5=0$

Solve for lambda by using the quadratic formula.

$\lambda_{1,2}=1\pm2i$

Find the eigenvalues of the following matrix, if possible.

$A=\begin{bmatrix} 1&2 \ -2&1 \end{bmatrix}$

$\lambda_{1,2}=1\pm2i$

$\lambda_{1,2}=\pm1$

$\lambda=\frac{5}{2}$

$\lambda=\begin{bmatrix} 2i&4i \ -2i&i \end{bmatrix}$

The eigenvalues do not exist.

Explanation

In order to find the eigenvalues of a matrix, apply the following formula:

$det(A-\lambda I) =0$

is the identity matrix.

$\begin{bmatrix} 1&2 \ -2&1 \end{bmatrix}-\lambda \begin{bmatrix} 1&0 \ 0&1 \end{bmatrix}=\begin{bmatrix} 1-\lambda & 2\ -2&1-\lambda \end{bmatrix}$

Compute the determinant and set it equal to zero.

$(1-\lambda)(1-\lambda)-(2)(-2)=0$

$1-2\lambda+\lambda^2+4=0$

$\lambda^2-2\lambda+5=0$

Solve for lambda by using the quadratic formula.

$\lambda_{1,2}=1\pm2i$

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