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Math  /  Algebra

QuestionFind the characteristic equation, the eigenvalues, and bases for the eigenspaces of the following matrix: A=[84−110−3010−4]A=\left[\begin{array}{ccc} 8 & 4 & -11 \\ 0 & -3 & 0 \\ 1 & 0 & -4 \end{array}\right]
The characteristic equation is □\square =0=0 NOTE: Enter the eigenvalues in increasing order.
Eigenvalues λ=\lambda= □\square λ=\lambda= □\square □\square Choose one Choose one -

Studdy Solution

STEP 1

What is this asking? We need to find a special equation called the characteristic equation, then use it to find special values called eigenvalues, and finally, find the eigenvectors that go with each eigenvalue! Watch out! Don't mix up the rows and columns of the matrix when calculating the determinant – that's a classic slip-up!
Also, remember that eigenvectors are not unique; any non-zero scalar multiple of an eigenvector is also an eigenvector.

STEP 2

1. Find the characteristic equation
2. Calculate the eigenvalues
3. Determine the eigenspaces

STEP 3

To **find** the characteristic equation, we need to solve ∣A−λI∣=0|A - \lambda I| = 0.
Here, λ\lambda represents our **eigenvalues** and II is the **identity matrix**.
This equation is super important because it helps us find those special eigenvalues that make the matrix AA act like a simple scalar multiplication.

STEP 4

Let's **calculate** the determinant! ∣8−λ4−110−3−λ010−4−λ∣=(8−λ)∣−3−λ00−4−λ∣−11∣0−3−λ10∣\begin{vmatrix} 8-\lambda & 4 & -11 \\ 0 & -3-\lambda & 0 \\ 1 & 0 & -4-\lambda \end{vmatrix} = (8-\lambda)\begin{vmatrix} -3-\lambda & 0 \\ 0 & -4-\lambda \end{vmatrix} - 11\begin{vmatrix} 0 & -3-\lambda \\ 1 & 0 \end{vmatrix} =(8−λ)(−3−λ)(−4−λ)−4(0)−11(3+λ)= (8-\lambda)(-3-\lambda)(-4-\lambda) - 4(0) - 11(3+\lambda)=(8−λ)(−3−λ)(−4−λ)−33−11λ= (8-\lambda)(-3-\lambda)(-4-\lambda) - 33 - 11\lambda=(8−λ)(12+7λ+λ2)−33−11λ= (8-\lambda)(12 + 7\lambda + \lambda^2) - 33 - 11\lambda=96+56λ+8λ2−12λ−7λ2−λ3−33−11λ= 96 + 56\lambda + 8\lambda^2 - 12\lambda - 7\lambda^2 - \lambda^3 - 33 - 11\lambda=−λ3+λ2+33λ+63= -\lambda^3 + \lambda^2 + 33\lambda + 63

STEP 5

So, our **characteristic equation** is −λ3+λ2+33λ+63=0-\lambda^3 + \lambda^2 + 33\lambda + 63 = 0.
Multiplying by -1 gives us λ3−λ2−33λ−63=0\lambda^3 - \lambda^2 - 33\lambda - 63 = 0.

STEP 6

Now, we **solve** for λ\lambda to find our **eigenvalues**.
By trying out small integer values, we find that λ=−3\lambda = -3 is a root.
We can then perform polynomial division to get (λ+3)(λ2−4λ−21)=0(\lambda+3)(\lambda^2 - 4\lambda - 21) = 0.
Factoring the quadratic gives us (λ+3)(λ−7)(λ+3)=0(\lambda+3)(\lambda-7)(\lambda+3) = 0, or (λ+3)2(λ−7)=0(\lambda+3)^2(\lambda-7) = 0.

STEP 7

This gives us our **eigenvalues**: λ1=−3\lambda_1 = -3 (with multiplicity 2) and λ2=7\lambda_2 = 7.

STEP 8

For λ=−3\lambda = -3, we solve (A−(−3)I)v=0(A - (-3)I)v = 0, or (A+3I)v=0(A+3I)v = 0. [114−1100010−1][xyz]=[000]\begin{bmatrix} 11 & 4 & -11 \\ 0 & 0 & 0 \\ 1 & 0 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} From the first and third rows, we have 11x+4y−11z=011x + 4y - 11z = 0 and x−z=0x - z = 0, which means z=xz = x.
Substituting into the first equation, we get 11x+4y−11x=011x + 4y - 11x = 0, so 4y=04y = 0 and y=0y = 0.
Thus, our **eigenvector** is any non-zero scalar multiple of [101]\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}.

STEP 9

For λ=7\lambda = 7, we solve (A−7I)v=0(A - 7I)v = 0. [14−110−10010−11][xyz]=[000]\begin{bmatrix} 1 & 4 & -11 \\ 0 & -10 & 0 \\ 1 & 0 & -11 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} From the second row, we have −10y=0-10y = 0, so y=0y = 0.
From the first and third rows, we have x−11z=0x - 11z = 0, so x=11zx = 11z.
Thus, our **eigenvector** is any non-zero scalar multiple of [1101]\begin{bmatrix} 11 \\ 0 \\ 1 \end{bmatrix}.

STEP 10

The characteristic equation is λ3−λ2−33λ−63=0\lambda^3 - \lambda^2 - 33\lambda - 63 = 0.
The eigenvalues are λ1=−3\lambda_1 = -3, λ2=−3\lambda_2 = -3, and λ3=7\lambda_3 = 7.
The eigenspace for λ=−3\lambda = -3 is spanned by [101]\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} and the eigenspace for λ=7\lambda = 7 is spanned by [1101]\begin{bmatrix} 11 \\ 0 \\ 1 \end{bmatrix}.

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