Two Eigenvectors Corresponding to Distinct Eigenvalues are Linearly Independent

Linear algebra problems and solutions

Problem 187

Let $A$ be an $n\times n$ matrix. Suppose that $\lambda_1, \lambda_2$ are distinct eigenvalues of the matrix $A$ and let $\mathbf{v}_1, \mathbf{v}_2$ be eigenvectors corresponding to $\lambda_1, \lambda_2$, respectively.

Show that the vectors $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent.

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To show that the vectors $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent, consider a linear combination
\[c_1 \mathbf{v}_1+c_2\mathbf{v}_2=\mathbf{0}, \tag{*}\] where $c_1, c_2$ are complex numbers.
Our goal is to show that $c_1=c_2=0$.

By the definitions of eigenvalues and eigenvectors we have
\[A\mathbf{v}_1=\lambda_1 \mathbf{v}_1 \text{ and } A\mathbf{v}_2 =\lambda_2 \mathbf{v}_2.\] Multiplying $A$ and (*), we have
\mathbf{0}&=A\cdot \mathbf{0}\\
&=A(c_1 \mathbf{v}_1+c_2\mathbf{v}_2)\\
&=c_1 A\mathbf{v}_1+c_2A\mathbf{v}_2\\
&=c_1\lambda_1 \mathbf{v}_1+c_2\lambda_2\mathbf{v}_2. \tag{**}

Now we multiply $\lambda_2$ and (*) and we obtain
\[\mathbf{0}=c_1\lambda_2\mathbf{v}_1+c_2\lambda_2\mathbf{v}_2.\] We subtract (**) from the last expression, and get
-(c_1\lambda_1 \mathbf{v}_1+c_2\lambda_2\mathbf{v}_2)\\

Recall that an eigenvector is by definition a nonzero vector, and hence $\mathbf{v}_1\neq \mathbf{0}$.
Thus we must have
\[c_1(\lambda_2-\lambda_1)=0.\] Since $\lambda_1$ and $\lambda_2$ are distinct, we must have $c_1=0$.

Substituting $c_1=0$ into (*), we also see that $c_2=0$ since $\mathbf{v}_2\neq \mathbf{0}$.
Therefore, the values of $c_1$ and $c_2$ are both zero, and hence the eigenvectors $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent.

Related Question.

As an application of this problem, try the next problem, which provides a geometrical meaning of eigenvalues.

Problem. Let $T:\R^2 \to \R^2$ be a linear transformation and let $A$ be the matrix representation of $T$ with respect to the standard basis of $\R^2$.

Prove that the following two statements are equivalent.

  1. There are exactly two distinct lines $L_1, L_2$ in $\R^2$ passing through the origin that are mapped onto themselves:
    \[T(L_1)=L_1 \text{ and } T(L_2)=L_2.\]
  2. The matrix $A$ has two distinct nonzero real eigenvalues.

For a proof of this problem, see the post “A Linear Transformation Preserves Exactly Two Lines If and Only If There are Two Real Non-Zero Eigenvalues“.

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3 Responses

  1. Alan says:

    There is a typo on the first line of the proof. “To show that the vectors v1,v2 are linearly dependent” should say independent.

  1. 06/23/2017

    […] general eigenvectors corresponding to distinct eigenvalues are linearly independent. Thus, $mathbf{v}_1, mathbf{v}_2$ are linearly independent. Hence the lines $L_1, L_2$ spanned by […]

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