Question

Let $egin{aligned} A = egin{bmatrix} 5 & 4 & 2\ 4 & 5 & 2\ 2 & 2 & 2 end{bmatrix} . end{aligned}$ The characteristic polynomial of A is $p_A(lambda) = -(lambda - 1)^2(lambda - 10)$. (a) Is $egin{bmatrix} 2\ 2\ 1 end{bmatrix}$ an eigenvector of A? If so, what is the corresponding eigenvalue? (b) Find bases for the eigenspaces $ ext{null}(A - 10I)$ and $ ext{null}(A - I)$. (c) Is A diagonalizable? If so, give the matrices D and P so that $A = PDP^{-1}$.

          Let
$egin{aligned} A = egin{bmatrix} 5 & 4 & 2\ 4 & 5 & 2\ 2 & 2 & 2 end{bmatrix} . end{aligned}$
The characteristic polynomial of A is $p_A(lambda) = -(lambda - 1)^2(lambda - 10)$.
(a) Is $egin{bmatrix} 2\ 2\ 1 end{bmatrix}$ an eigenvector of A? If so, what is the corresponding eigenvalue?
(b) Find bases for the eigenspaces $	ext{null}(A - 10I)$ and $	ext{null}(A - I)$.
(c) Is A diagonalizable? If so, give the matrices D and P so that $A = PDP^{-1}$.

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Let
eginaligned A = eginbmatrix 5 4 2 4 5 2 2 2 2 endbmatrix . endaligned
The characteristic polynomial of A is pA(lambda) = -(lambda - 1)^2(lambda - 10).
(a) Is eginbmatrix 2 2 1 endbmatrix an eigenvector of A? If so, what is the corresponding eigenvalue?
(b) Find bases for the eigenspaces extnull(A - 10I) and extnull(A - I).
(c) Is A diagonalizable? If so, give the matrices D and P so that A = PDP^-1.

Added by Collin R.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Krishna G

Step 1

The eigenvalues are the roots of the characteristic polynomial, which is given by: $P_A(A) = -(A - 1)^2(A - 10)$ The roots of this polynomial are the eigenvalues of A, which are 1 and 10. Now, let's find the eigenvectors corresponding to these Show more…

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Let A = 4. The characteristic polynomial of A is PA(A) = -(A - 1)^2(A - 10). Is there an eigenvector of A? If so, what is the corresponding eigenvalue? (b) Find bases for the eigenspaces null(A - 10I) and null(A - I). Is A diagonalizable? If so, give the matrices P and D so that PDP^-1.

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