# Matrices

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Let B = . Find each of the following. You may use a calculator to replace hand calculation for row reduction and finding the characteristic polynomial. Show your work; don't just write down the answers.

(a) Find the characteristic polynomial of "A".

(b) Find the eigenvalues of "A"; for each, find the algebraic multiplicity.

(c) Find a basis for each eigenspace.

(d) Diagonalize "A" ; that is, find: S , Λ , and, S-1 such that A = S Λ S-1 .

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Let B = . Find each of the following. You may use a calculator to replace hand calculation for row reduction and finding the characteristic polynomial. Show your work; don't just write down the answers.

(a) Find the characteristic polynomial of "A".

(b) Find the eigenvalues of "A"; for each, find the algebraic multicplicity.

(c) Find a basis for each eigenspace.

(d) Diagonalize "A" ; that is, find: S , Î› , and, S-1 such that A = S Î› S-1 .

Solution:

A is a block triangular with diagonal blocks [1] , and [1].

Let P =

= (t - 1)2 - 4 = t2- 2t - 3 = (t - 3)(t + 1)

The characteristic polynomial is

= (t - 1)2(t - 3)(t+1)

Solution (b):

The eigen values of A are t1 = 1 , t2 = 3 , t3 = - 1

Since (t - 1) occurs twice in the characteristic polynomial, the algebraic multiplicity of t1 is two.

The algebraic multiplicity of t2 is one, and the algebraic multiplicity of t3 is also one.

Solution (c)

Now we shall find the eigen space for the eigen value t1 = 1

ïƒž

By R4 - R3 we get

ïƒž y + 2z + 2w = 0

x + 2z + 2w = 0

2x +2y + w = 0

z - w = 0

z - w = 0 ïƒž z = w

Put z = w = 1, then we get y + 2(1) + 2(1) = 0 ïƒž y = - 4

And x + 2(1) + 2(1) = 0 ïƒž x = - 4

Therefore, one eigen vector is ( - 4 , - 4 , 1 , 1 )

Similarly, put z = - 1 , w = - 1 , we get (4, 4 , - 1 , - 1 )

Therefore, the eigen space corresponding to t1 = 1

is {( - 4 , - 4 , 1 , 1 ) , (4, 4 , - 1 , - 1 ) }

Now we find the eigen space corresponding to t2 = 3.

By R4 - R3 , we get the following

By R3 + R1 we get the following

- 2x + y + 2z + 2w = 0

x - 2y + 2z + 2w = 0

3y + 3w = 0

3z - 3w = 0

Now 3z - 3w = 0 ïƒž z = w

Again 3y + 3w = 0 ïƒž y = - w

Substituting in x - 2y + 2z + 2w = 0 we get x + 2w + 2w + 2w = 0

ïƒž x = - 6w

Therefore , the eigen space is ( - 6 , - 1 , 1 , 1 )

Now finally, we find the eigen space corresponding to t3 = - 1

ïƒž

By R4 - R3 we get the following

Further, by R3 - R1 we get the following

2x + y + 2z + 2w = 0

x + 2y + 2z + 2w = 0

y - w = 0

- z + w = 0

- z + w = 0 ïƒž z = w

Again y - w = 0 ïƒž y = w

From the above, we get z = - 6 , y = 1 , z = 1 , w = 1

Therefore, the eigen space corresponding to t3 = - 1 is (- 6,1,1,1)

Solution (d):

Writing the above four eigen spaces as four columns, we get the required matrix S which transforms A to diagonal matrix.

Therefore,

such that SAS-1 = A

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