Matrix A has characteristic polynomial p(𝜆) = (−45 − 𝜆)(5 − 𝜆)(27 − 𝜆).A = −3 4 32 4 3 −1632 −16 −13Given this information, find the spectral form of A, i.e. A = QΛQT

To find the spectral form of matrix A, we need to find the eigenvalues and eigenvectors of A.

From the given characteristic polynomial p(𝜆) = (−45 − 𝜆)(5 − 𝜆)(27 − 𝜆), we can determine the eigenvalues of A. The eigenvalues are the values of 𝜆 that satisfy p(𝜆) = 0.

Setting each factor of the characteristic polynomial equal to zero, we have: -45 - 𝜆 = 0, which gives 𝜆 = -45 5 - 𝜆 = 0, which gives 𝜆 = 5 27 - 𝜆 = 0, which gives 𝜆 = 27

So, the eigenvalues of A are 𝜆 = -45, 5, and 27.

To find the eigenvectors corresponding to each eigenvalue, we substitute each eigenvalue back into the equation (A - 𝜆I)x = 0, where I is the identity matrix.

For 𝜆 = -45: (A - (-45)I)x = 0 (-3 - (-45))(4 32) (4 3 - (-45))(−16 −13)x = 0

Simplifying the equation, we get: 48x1 + 32x2 = 0 4x1 + 48x2 = 0

Solving this system of equations, we find that x1 = -2x2. So, the eigenvector corresponding to 𝜆 = -45 is [-2, 1].

Similarly, for 𝜆 = 5 and 𝜆 = 27, we can find the corresponding eigenvectors.

For 𝜆 = 5: (A - 5I)x = 0 (-3 - 5)(4 32) (4 3 - 5)(−16 −13)x = 0

Simplifying the equation, we get: -8x1 + 32x2 = 0 4x1 - 2x2 = 0

Solving this system of equations, we find that x1 = 4x2. So, the eigenvector corresponding to 𝜆 = 5 is [4, 1].

For 𝜆 = 27: (A - 27I)x = 0 (-3 - 27)(4 32) (4 3 - 27)(−16 −13)x = 0

Simplifying the equation, we get: -30x1 + 32x2 = 0 4x1 - 30x2 = 0

Solving this system of equations, we find that x1 = 8x2. So, the eigenvector corresponding to 𝜆 = 27 is [8, 1].

Now, we can form the matrix Q using the eigenvectors as columns: Q = [-2 4 8 1 1 1]

To find the diagonal matrix Λ, we place the eigenvalues on the diagonal: Λ = [-45 0 0 0 5 0 0 0 27]

Finally, we can write the spectral form of matrix A as A = QΛQT: A = [-2 4 8] [-45 0 0] [-2 1] [1 1 1] [0 5 0] [4 1] [0 0 27]