Find the eigenvalues and a basis for an eigenspace of matrix A

Question

Find the eigenvalues and a basis for each eigenspace of matrix A:

\begin{bmatrix} 1 & -3 & 3 \\ 2 & -2 & 2 \\ 2 & 0 & 0 \\ \end{bmatrix}

I found the eigenvalues by computing $|A-\lambda I|$:

$\lambda_1 = 0,$ $\lambda_2 = 1,$ $\lambda_3 = -2$

How do I find a basis for each eigenspace of matrix A?

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I tried the following:

$\lambda = 0:$

\begin{bmatrix} 1 & -3 & 3 \\ 2 & -2 & 2 \\ 2 & 0 & 0 \\ \end{bmatrix}

Do reduced row echelon form:

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ \end{bmatrix}

Does this mean one of the results is \begin{bmatrix} 0\\ t\\ t\\ \end{bmatrix}

and to find the other two answers, do the same thing except set $\lambda$ equal to the other two values? Or does finding the basis for each eigenspace of matrix A mean something different?

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Answer 1

Correct, you subtract each eigenvalue from the main diagonal and use row reduction to find the eigenvectors. You may also want to spend a minute or two trying to spot the eigenvectors because that tends to be faster than row reduction. For instance in your case the second and third columns are scalar multiples of each other and you can use this information to get the eigenvector $(0,1,1)^T$, which corresponds to the eigenspace $\{(0,t,t)^T : t \in \mathbf{R} \}$.

For the eigenvalue of $1$ you are looking for a vector $v$ with $Av = v$. If $v = (a, b, c)^T$ then $Av = (a - 3b + 3c, 2a - 2b + 2c, 2a)^T$. Thus $2a = c$ and we can now do this again with $A(a,b,2a)^T = (7a - 3b, 6a - 2b, 2a)^T$. This gives you the equations $7a - 3b = a$ and $6a - 2b = b$, both equivalent to $6a - 3b = 0$. Hence one eigenvector ($a = 1$) is $(1,2,2)^T$. This process is basically row reduction. I am making use of the fact that the third row has just one entry to try to move through the steps a bit faster. For practice, use row reduction to verify that this is the correct eigenvector.

Answer 2

You’ve described the general process of finding bases for the eigenspaces correctly. Note that since there are three distinct eigenvalues, each eigenspace will be one-dimensional (i.e., each eigenspace will have exactly one eigenvector in your example). If there were less than three distinct eigenvalues (e.g. $\lambda$=2,0,2 or $\lambda$=2,1), there would be at least one eigenvalue that yields more than one eigenvector. For this matrix, however, you can find the eigenvalues ($\lambda$s) and at least two corresponding eigenvectors by inspection.

Observe that the product of a matrix and column vector is a linear combination of the columns of the matrix with coefficients given by the vector. Adding the 2nd column to the 3rd produces 0 (the column vector of all 0s), so $\begin{bmatrix} 0\\1\\1 \end{bmatrix}$ is an eigenvector of $\lambda=0$. Because these two columns cancel each other if their coefficients in a linear combination are equal, one can also see that $A\begin{bmatrix} 1\\2\\2 \end{bmatrix} = \begin{bmatrix} 1\\2\\2 \end{bmatrix}$, so $\begin{bmatrix} 1\\2\\2 \end{bmatrix}$ is an eigenvector of $\lambda=1$. You get the third eigenvalue “for free” because the trace of the matrix is equal to the sum of the eigenvalues, so the last eigenvalue is $1+(-2)-(0+1)=-2$. An eigenvector for this last eigenvalue might be harder to spot than the first two (subtract the 1st column from the 3rd), so you can always fall back on the standard method for this last eigenspace, namely, finding the null space of $A+2I$.