Question

1.

Find the sum and product of the eigenvalues of the matrix {1 1 1}{1 2 2 } { 1 2 3}​

Answer 1

Answer:

Step-by-step explanation:

sum of given matrix

1. (1+1+1  1+2+2 1+2+3)
2. (3 5 6)

product

1. (1*1*1 1*2*2 1*2*3)
2. (1 4 6)

eigenvalues

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Answer 2

Answer:

This proof requires the investigation of the characteristic polynomial of AA, which is found by taking the determinant of (A - \lambda{I}_{n})(A−λI

n

).

A = \begin{bmatrix} {a}_{11} & \cdots &{a}_{1n} \\ \vdots &\ddots &\vdots \\ {a}_{n1} &\cdots & {a}_{nn}\\ \end{bmatrix}

A=

⎣

⎢

⎡

a

11

⋮

a

n1

⋯

⋱

⋯

a

1n

⋮

a

nn

⎦

⎥

⎤

A - {I}_{n}\lambda = \begin{bmatrix} {a}_{11} - \lambda & \cdots &{a}_{1n} \\ \vdots &\ddots &\vdots \\ {a}_{n1} &\cdots & {a}_{nn} - \lambda\\ \end{bmatrix}

A−I

n

λ=

⎣

⎢

⎡

a

11

−λ

⋮

a

n1

⋯

⋱

⋯

a

1n

⋮

a

nn

−λ

⎦

⎥

⎤

Observe that det(A - \lambda{I}_{n} ) = det(A) + ... + tr(A){(-\lambda)}^{n-1} + {(-\lambda)}^{n}det(A−λI

n

)=det(A)+...+tr(A)(−λ)

n−1

+(−λ)

n

.

Let {r}_{1}, {r}_{2}, ...,{r}_{n}r

1

,r

2

,...,r

n

be the roots of an n-order polynomial.

P(\lambda) = ({r}_{1} - \lambda)({r}_{2} - \lambda)...({r}_{n} - \lambda)

P(λ)=(r

1

−λ)(r

2

−λ)...(r

n

−λ)

P(\lambda) = \prod _{ i=i }^{ n }{ { r }_{ i } } +...+\sum _{ i=i }^{ n }{ { r }_{ i }{(-\lambda)}^{n-1} + {(-\lambda)}^{n} }

P(λ)=

i=i

∏

n

r

i

+...+

i=i

∑

n

r

i

(−λ)

n−1

+(−λ)

n

Since the eigenvalues are the roots of a matrix polynomial, we can match P(x)P(x) to det(A - \lambda{I}_{n})det(A−λI

n

). Therefore it is clear that

\prod _{ i=i }^{ n }{ { \lambda }_{ i } = det(A)}

i=i

∏

n

λ

i

=det(A)

and

\sum _{ i=i }^{ n }{ { \lambda }_{ i } = tr(A)}.

i=i

∑

n

λ

i

=tr(A).