Linear Algebra. Help please.

[thomasween]

With respect to the linear operator T (x, y, z, t) = (3x, x + 2y + 3z + t, 4x + z, 5x + 2z) of R⁴.

I. Is the characteristic polynomial of T p (x) = x⁴ − 6x³ + 11x² − 6x?

II. Are the eigenvalues associated with T 0,1,1,3?

III. T is diagonalizable?

 

[Deleted member 4993]

With respect to the linear operator T (x, y, z, t) = (3x, x + 2y + 3z + t, 4x + z, 5x + 2z) of R⁴.

I. Is the characteristic polynomial of T p (x) = x⁴ − 6x³ + 11x² − 6x?

II. Are the eigenvalues associated with T 0,1,1,3?

III. T is diagonalizable?

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem.

 

[HallsofIvy]

This linear operator can be represented by the matrix \(\displaystyle \begin{bmatrix}3 & 0 & 0 & 0 \\ 1 & 2 & 3 & 1 \\ 4 & 0 & 1 & 0 \\ 5 & 0 & 2 & 0 \end{bmatrix}\).

It's characteistic equation is \(\displaystyle \left|\begin{array}{cccc}{3-\lambda} & 0 & 0 & 0 \\ 1 & 2-\lambda & 3 & 1 \\ 4 & 0 & 1-\lambda & 0 \\ 5 & 0 & 2 & -\lambda \end{array}\right|= 0\). The characteristic polynomial is the left side of that.

 

[Deleted member 4993]

This linear operator can be represented by the matrix \(\displaystyle \begin{bmatrix}3 & 0 & 0 & 0 \\ 1 & 2 & 3 & 1 \\ 4 & 0 & 1 & 0 \\ 5 & 0 & 2 & 0 \end{bmatrix}\).

It's characteistic equation is \(\displaystyle \left|\begin{array}{cccc}{3-\lambda} & 0 & 0 & 0 \\ 1 & 2-\lambda & 3 & 1 \\ 4 & 0 & 1-\lambda & 0 \\ 5 & 0 & 2 & -\lambda \end{array}\right|= 0\). The characteristic polynomial is the left side of that.

~5 days went by after the request to show work - OP is AWOL!!

 

[HallsofIvy]

May he rest in peace.