For the following square matrix:

$$ \left( \begin{array}{ccc} 3 & 0 & 1 \\ -4 & 1 & 2 \\ -6 & 0 & -2 \end{array} \right)$$

Decide which, if any, of the following vectors are eigenvectors of that matrix and give the corresponding eigenvalue.

$ \left( \begin{array}{ccc} 2 \\ 2 \\ -1 \end{array} \right)$ $ \left( \begin{array}{ccc} -1 \\ 0 \\ 2 \end{array} \right)$ $ \left( \begin{array}{ccc} -1 \\ 1 \\ 3 \end{array} \right)$ $ \left( \begin{array}{ccc} 0 \\ 1 \\ 0 \end{array} \right)$$ \left( \begin{array}{ccc} 3 \\ 2 \\ 1 \end{array} \right)$

If I've understood correctly, I must multiply the matrix by each vector first. If the result is a multiple of that vector, then it's an eigenvector. Only the fourth vector is so. But how should I calculate its corresponding eigenvalue?


Gigili, this is how's usually calculated both eigenvalues and eigenvectors. Calculate the determinant:

$$\det(\lambda I-A)=\begin{vmatrix}\lambda-3&0&-1\\4&\lambda-1&-2\\6&0&\lambda+2\end{vmatrix}=(\lambda-1)(\lambda-3)(\lambda+2)+6(\lambda-1)=\lambda(\lambda-1)^2\Longrightarrow$$

$$\Longrightarrow \lambda=0\,,\,1\,\,\,\text{are the eigenvalues of the matrix}$$

Now, to find eigenvectors corresponding to the eigenvalues you form a homogeneous linear system by subtituting $\,\lambda\,$ in the above matrix expression with the corr. value. Call the unknowns $\,x,y,z\,$ and note that since the determinant is going to be zero we get always a system with a non-trivial expression (why?):


Thus, any eigenvector corresponding to the eigenvalue $\,\lambda=0\,$ has the form

$$\begin{pmatrix}\;\;x\\\!\!-2x\\\!\!-3x\end{pmatrix}\,\,,\,\,\text{for example}\,\,\,\begin{pmatrix}\;\;\;1\\-2\\-3\end{pmatrix}$$

Try now to mimic the above for the eigenvalue $\,\lambda=1\,$ . Once again, you get a system of rank 2 and thus there's only one linearly independent eigenvector, too.

  • $\begingroup$ Thank you, that was really helpful. For $\lambda=1$, I've got $z=-2x$ and nothing for $y$. Could you elaborate a bit? $\endgroup$ – Gigili Jan 21 '13 at 11:18
  • $\begingroup$ That means the eigenvectors of $\,\lambda =1\,$ look like $$\begin{pmatrix}x\\y\\-2x\end{pmatrix}$$ . Choose now wisely values for the free parameters $\,x,y\,$ and get two linearly independent eigenvectors. This cannot be done anymore elaborated in this means: you can read it in any decent linear algebra book, the web...or wait until your teacher reaches the subject. $\endgroup$ – DonAntonio Jan 21 '13 at 12:17
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    $\begingroup$ No, @Gigili: i was wrong. When substituting $\,\lambda=1\,$ you do not get two lin. indep. eigenvectors but only one: the first and second equations you get are linearly independent so the system's rank is two, not one as I wrote. Solving this system we actually get $\,x=z=0\,\,,\,\,y$ anything different from zero, for example $\,\begin{pmatrix}0\\1\\0\end{pmatrix}$ , for example the 4th vector. $\endgroup$ – DonAntonio Jan 21 '13 at 12:47
  • $\begingroup$ I shall edit my answer now to correct this. $\endgroup$ – DonAntonio Jan 21 '13 at 12:47

If $\mathbf{M}\mathbf{v} = \lambda \mathbf{v}$ for a scalar $\lambda$, then $\mathbf{v}$ is an eigenvector of $\mathbf{M}$, and $\lambda$ is its corresponding eigenvalue.

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    $\begingroup$ Thank you for your answer. So in the above problem, $\lambda=1$, am I right? $\endgroup$ – Gigili Jan 21 '13 at 8:24
  • $\begingroup$ No, that's not correct. What do you get for $\mathbf{M}\mathbf{v}$? $\endgroup$ – user7530 Jan 21 '13 at 8:25
  • $\begingroup$ $ \left( \begin{array}{ccc} 0 \\ 1 \\ 0 \end{array} \right)$ is what I've got as for the 4th vector, which appears to be an eigenvector. Isn't that correct? $\endgroup$ – Gigili Jan 21 '13 at 8:28
  • $\begingroup$ Oh, for some reason I thought you'd written $\lambda = -1$. $\lambda = 1$ is correct. $\endgroup$ – user7530 Jan 21 '13 at 8:29
  • $\begingroup$ Oh sorry, I edited it while you were responding to my comment! Thank you anyway. $\endgroup$ – Gigili Jan 21 '13 at 8:30

To expand on DonAntonio's answer: an eigenvalue and eigenvector of a $n \times n$ matrix $A$ is defined as a respectively a scalar $\lambda$ and a vector $x \neq 0$ such that $Ax = \lambda x$. But this equivalent to $Ax - \lambda x = (A-\lambda I)x = 0$. We now want to solve this equation for $x$, so that $x \neq 0$. Now, if $A-\lambda I$ would be invertible, we would have only the solution $x = (A-\lambda I)^{-1} 0 = 0$, so that's not what we want. Instead, we require the matrix $A-\lambda I$ to be singular.

Recall that a matrix is singular if and only if it's determinant is zero, so we calculate the determinant of $A - \lambda I$ and equate it to zero: $$ 0 = det(A-\lambda I) = (\lambda - \lambda_1)^{m_1} \ldots (\lambda - \lambda_k)^{m_k} $$ This is called the characteristic equation for $A$. This will yield a polynomial of degree $n$, with $k \leq n$ roots. These roots are then the eigenvalues of $A$, since these values of $\lambda$ will make $A-\lambda I$ singular. The exponentials $m_1$ through $m_k$ are called the algebraic multiplicities of the corresponding eigenvalues by the way, and these determine an upper bound on the dimensions of the eigenspaces, that I will explain next.

So now, we have a matrix $A - \lambda_i I$ that is singular, and we want to solve $(A-\lambda_i I)x_i = 0$. Every vector $x_i$ that is a solution of this equation is an eigenvector of $A$, corresponding to the eigenvalue $\lambda_i$. But, this vector is not unique! Note that if $x_i$ is an eigenvector, then so is $c x_i$ for every scalar $c \neq 0$. And it could even be the case that there exist two or more linearly independent eigenvectors! The space of all vectors that are a solution to this equation are called the eigenspace of $A$ for the eigenvector $\lambda_i$. This space is the same as the null space of $A-\lambda_i I$, whose dimension is determined by the rank-nullity theorem. This theorem (look it up in your linear algebra book ;) ) states that the dimension of a matrix is equal to the sum of it's rank (the dimension of the row space) and it's nullity (the dimension of it's null space). It also can be shown that the dimension of the eigenspace must be necessarily less or equal to the algebraic multiplicity of the eigenvector. This dimension is called the geometric multiplicity of $\lambda_i$.

So, to summarize the calculation of eigenvalues and corresponding eigenvectors:

  • Write down the characteristic polynomial of $A$: $$ det(A-\lambda I) = 0. $$
  • Solve the characteristic equation. The solutions $\lambda_i$ are the eigenvalues of $A$.
  • Write down the system $(A-\lambda I) x = 0$ and solve the system for the vector $x$. (By Gaussian elimination or something like that.) The solutions $x_i$ are the eigenvectors of $A$ corresponding to $\lambda_i$. The subspace of all eigenvectors corresponding to $\lambda_i$ is the eigenspace of $A$ for $\lambda_i$.

Hope that helped a bit. ;)

  • $\begingroup$ This is a great add-on, thank you. $\endgroup$ – Gigili Jan 21 '13 at 12:37

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