(1 point) Supppose A is an invertible n×n matrix and v⃗ is an eigenvector of A with associated eigenvalue 8. Convince yourself that v⃗ is an eigenvector of the following matrices, and find the associated eigenvalues. The matrix A4 has an eigenvalue . The matrix A−1 has an eigenvalue . The matrix A+4In has an eigenvalue . The matrix 5A has an eigenvalue .

Question

20-04-2020
Mathematics

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(1 point) Supppose A is an invertible n×n matrix and v⃗ is an eigenvector of A with associated eigenvalue 8. Convince yourself that v⃗ is an eigenvector of the following matrices, and find the associated eigenvalues. The matrix A4 has an eigenvalue . The matrix A−1 has an eigenvalue . The matrix A+4In has an eigenvalue . The matrix 5A has an eigenvalue .

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LammettHash LammettHash · Answer 1 · 2020-04-20T21:08:15+02:00

By definition, if [tex]\vec v[/tex] is an eigenvector of [tex]\mathbf A[/tex] (with associated eigenvalue 8), then

[tex]\mathbf A\vec v=8\vec v[/tex]

Notice that

[tex]\mathbf A^4\vec v=\mathbf A^3\cdot\mathbf A\vec v=8\mathbf A^3\vec v[/tex]

Rinse and repeat to find

[tex]\mathbf A^4\vec v=8^4\vec v[/tex]

so that [tex]\mathbf A^4[/tex] has a corresponding eigenvalue of 4096.

Let [tex]\lambda[/tex] be the eigenvalue of [tex]\mathbf A^{-1}[/tex] corresponding to [tex]\vec v[/tex]. Then

[tex]\mathbf A^{-1}\vec v=\lambda\vec v\implies\mathbf A\cdot\mathbf A^{-1}\vec v=\mathbf A\cdot\lambda \vec v\implies\vec v=8\lambda\vec v[/tex]

For this to hold, we require [tex]8\lambda=1[/tex], or [tex]\lambda=\frac18[/tex]. So [tex]\mathbf A^{-1}[/tex] has a corresonding eigenvalue of 1/8.

Expanding gives

[tex](\mathbf A+4\mathbf I_n)\vec v=\mathbf A\vec v+4\mathbf I_n\vec v=8\vec v+4\vec v=12\vec v[/tex]

so that [tex]\mathbf A+4\mathbf I_n[/tex] has an associated eigenvalue of 12.

You know the drill:

[tex]5\mathbf A\vec v=5\cdot8\vec v=40\vec v[/tex]

so the eigenvalue of [tex]5\mathbf A[/tex] is 40.

ibiscollingwood ibiscollingwood · Answer 2 · 2022-04-08T08:32:36+02:00

The matrix [tex]A^{4}[/tex] has eigen value of 4096.

The matrix [tex]A^{-1}[/tex] has eigen value of [tex]\frac{1}{8}[/tex]

The eigen value of (A+4I) is 12.

The eigen value of [tex]5Av[/tex] is 40.

Eigen values of matrix:

Eigenvalues are the set of scalar values that is associated with the set of linear equations most probably in the matrix equations.

The eigenvectors are also termed as characteristic roots.

If v is an eigen vector of matrix A associated with eigen value [tex]\lambda[/tex].

Then, [tex]Av=\lambda v[/tex]

it is given that, v is an eigen vector of A with eign value 8.

so taht, [tex]Av=8v[/tex]

We have to find eigen value for [tex]A^{4}[/tex].

[tex]A^{4}v=A^{3}(8v)=8^{4}v=4096v[/tex]

Thus, [tex]A^{4}[/tex] has eigen value of 4096.

If A has eigen value of 8 . Then [tex]A^{-1}[/tex] has eigen value of [tex]\frac{1}{8}[/tex]

The eigen value of (A+4I) is computed as,

[tex](A+4I_{n})v=Av+4v=8v+4v=12v[/tex]

The eigen value of (A+4I) is 12.

The eigen value of [tex]5Av[/tex] is,

[tex]5Av=5*8v=40v[/tex]

The eigen value of [tex]5Av[/tex] is 40.

Learn more about the eigen vectors here:

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