Question

A 3 by 3 matrix Bis known to have eigenvalues 0, 1, 2. This information is enough to find three of these (give the answers where possible) : (a) the rank of B (b) thedeterminantofBTB (c) theeigenvaluesofBTB (d) the eigenvalues of (B2 + J)-1.

Answer 1

Answer with Step-by-step explanation:

We are given that a matrix B .

The eigenvalues of matrix are 0, 1 and 2.

a.We know that

Rank of matrix B=Number of different eigenvalues

We have three different eigenvalues

Therefore, rank of matrix B=3

b.

We know that

Determinant of matrix= Product of eigenvalues

Product of eigenvalues=
`0* 1* 2=0`

After transpose , the value of determinant remain same.

`\mid B^TB\mid=\mid B^T\mid \mid B\mid =0* 0=0`

c.Let

B=
`\left[\begin{array}{ccc}0&-&-\\-&1&-\\-&-&2\end{array}\right]`

Transpose of matrix:Rows change into columns or columns change into rows.

After transpose of matrix B

`B^T=\left[\begin{array}{ccc}0&-&-\\-&1&-\\-&-&2\end{array}\right]`

`B^TB=\left[\begin{array}{ccc}0^2&-&-\\-&1^2&-\\-&-&2^2\end{array}\right]`

`B^TB=\left[\begin{array}{ccc}0&-&-\\-&1&-\\-&-&4\end{array}\right]`

Hence, the eigenvalues of
`B^TB` are 0, 1 and 4.

d.Eigenvalue of Identity matrix are 1, 1 and 1.

Eigenvalues of
`B^2+I=(0+1),(1+1),(2^2+1)=1,2,5`

We know that if eigenvalue of A is
`\lambda`

Then , the eigenvalue of
`A^(-1)` is
`(1)/(\lambda)`

Therefore, the eigenvalues of
`(B^2+I)^(-1)` are

`(1)/(1),(1)/(2),(1)/(5)`

The eigenvalues of
`(B^2+I)^(-1)` are 1,
`(1)/(2)` and
`(1)/(5)`