Questions: Attempt 1:10 attempts remaining. Find the eigenvalues λ₁<λ₂ and associated unit eigenvectors u₁, u₂ of the symmetric matrix A=[ −7 −16 −16 17 ] The smaller eigenvalue λ₁= has associated unit eigenvector u₁=[ ]. The larger eigenvalue λ₂= has associated unit eigenvector u₂=[ ]. Note: The eigenvectors above form an orthonormal eigenbasis for the column space of A.

Attempt 1:10 attempts remaining.
Find the eigenvalues λ₁<λ₂ and associated unit eigenvectors u₁, u₂ of the symmetric matrix
A=[
−7 −16
−16 17
]

The smaller eigenvalue λ₁=  has associated unit eigenvector u₁=[

].

The larger eigenvalue λ₂=  has associated unit eigenvector u₂=[

].
Note: The eigenvectors above form an orthonormal eigenbasis for the column space of A.
Transcript text: Attempt 1:10 attempts remaining. Find the eigenvalues $\lambda_{1}<\lambda_{2}$ and associated unit eigenvectors $\vec{u}_{1}, \vec{u}_{2}$ of the symmetric matrix \[ A=\left[\begin{array}{cc} -7 & -16 \\ -16 & 17 \end{array}\right] \] The smaller eigenvalue $\lambda_{1}=$ $\square$ has associated unit eigenvector $\vec{u}_{1}=\left[\begin{array}{r}\square \\ \square \\ \square\end{array}\right]$. The larger eigenvalue $\lambda_{2}=$ $\square$ has associated unit eigenvector $\vec{u}_{2}=$ $\left[\begin{array}{l}i_{+} \\ \vdots\end{array}\right]$. Note: The eigenvectors above form an orthonormal eigenbasis for the column space of $A$. Submit answer Next item
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Solution

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Solution Steps

To find the eigenvalues and associated unit eigenvectors of a symmetric matrix, we first compute the eigenvalues by solving the characteristic equation, which is derived from the determinant of \(A - \lambda I = 0\). Once the eigenvalues are found, we substitute each back into the equation \((A - \lambda I)\vec{v} = 0\) to find the corresponding eigenvectors. Finally, we normalize these eigenvectors to obtain unit eigenvectors.

Step 1: Compute Eigenvalues

To find the eigenvalues of the matrix

\[ A = \begin{bmatrix} -7 & -16 \\ -16 & 17 \end{bmatrix}, \]

we solve the characteristic equation given by

\[ \det(A - \lambda I) = 0. \]

The eigenvalues obtained are

\[ \lambda_1 = -15 \quad \text{and} \quad \lambda_2 = 25. \]

Step 2: Compute Eigenvectors

Next, we find the eigenvectors associated with each eigenvalue. For \(\lambda_1 = -15\), we solve

\[ (A - (-15)I)\vec{u}_1 = 0, \]

which yields the eigenvector

\[ \vec{u}_1 = \begin{bmatrix} -0.8944 \\ -0.4472 \end{bmatrix}. \]

For \(\lambda_2 = 25\), we solve

\[ (A - 25I)\vec{u}_2 = 0, \]

resulting in the eigenvector

\[ \vec{u}_2 = \begin{bmatrix} -0.4472 \\ 0.8944 \end{bmatrix}. \]

Step 3: Normalize Eigenvectors

The eigenvectors \(\vec{u}_1\) and \(\vec{u}_2\) are already unit vectors, as they have been computed from the symmetric matrix \(A\).

Final Answer

The smaller eigenvalue is

\(\lambda_1 = -15\) with associated unit eigenvector

\(\vec{u}_1 = \begin{bmatrix} -0.8944 \\ -0.4472 \end{bmatrix}\).

The larger eigenvalue is

\(\lambda_2 = 25\) with associated unit eigenvector

\(\vec{u}_2 = \begin{bmatrix} -0.4472 \\ 0.8944 \end{bmatrix}\).

Thus, the final answers are:

\[ \boxed{\lambda_1 = -15, \quad \vec{u}_1 = \begin{bmatrix} -0.8944 \\ -0.4472 \end{bmatrix}, \quad \lambda_2 = 25, \quad \vec{u}_2 = \begin{bmatrix} -0.4472 \\ 0.8944 \end{bmatrix}} \]

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