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GRE Subject Test: Math : Eigenvalues

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Example Questions

Example Question #1 : Eigenvalues

Find the eigenvalues of the following matrix, if possible.

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

Possible Answers:

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

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

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

The eigenvalues do not exist.

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

Correct answer:

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

Explanation:

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

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

$\displaystyle I$ is the identity matrix.

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

Compute the determinant and set it equal to zero.

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

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

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

Solve for lambda by using the quadratic formula.

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

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Example Question #2 : Eigenvalues

Find the eigenvalues of the following matrix, if possible.

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

Possible Answers:

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

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

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

The eigenvalues do not exist.

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

Correct answer:

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

Explanation:

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

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

$\displaystyle I$ is the identity matrix.

Compute the determinant and set it equal to zero.

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

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

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

Solve for lambda by using the quadratic formula.

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

Report an Error

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GRE Subject Test: Math : Eigenvalues

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #1 : Eigenvalues

Example Question #2 : Eigenvalues

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