# What is the Relationship between the Rank and the Number of Eigenvalues?

for (a) it says it can't be rank 1 because there is only one non zero eigenvalue? I understand it can't be 0 or 3 but I don't understand how you can conclude we have 2 just by knowing how many eigenvalues we have.

• Eigenvectors corresponding to distinct eigenvalues are linearly independent. So if $x$ and $y$ are eigenvectors corresponding to eigenvalues $1$ and $2$, then $Bx = x$ and $By = 2y$ are also linearly independent, hence the dimension of the image (aka the rank) of $B$ is at least $2$. – Bungo Dec 13 '16 at 19:53
• Recall that the rank is the dimension of the image of $B$, where we view $B$ as a map from $V\rightarrow V$, $V$ being our underlying vector space. Eigenvectors of distinct eigenvalues are linearly independent. Can you show the result from here? – Krishan Bhalla Dec 13 '16 at 19:53
• @Bungo I must be missing something really obvious. I understand Bx = x By = 2y need to be linearly independent. But What does this say about B? – Hello Dec 13 '16 at 19:56
• $Bx$ and $By$ are elements of the image of $B$, and they are linearly independent. In general, if a vector space contains $k$ linearly independent vectors, its dimension must be at least $k$. (Here $k = 2$.) This is because any linearly independent set of vectors can be extended to a basis if it is not already a basis. – Bungo Dec 13 '16 at 19:58
• Yes, they are the same thing, just different names. (Similarly, kernel and null space mean the same thing.) – Bungo Dec 13 '16 at 20:01

Since $B$ is $3\times 3$, and has $3$ distinct eigenvalues, it is diagonalizable. So $B$ is similar to $$\begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$$

Since rank is a similarity invariant, it is now clear $B$ has rank $2$.

• This makes sense but I want to understand it more conceptually. Why does two disticnt nonzero eigenvalues = dimension two? – Hello Dec 13 '16 at 19:58
• @curiousBiggie If you take a basis of eigenvectors, the matrix acts by scaling these basis vectors. The image is then spanned by the image of the eigenvectors, and it's enough to take the span by the eigenvectors with nonzero eigenvalue, because those corresponding to eigenvalue $0$ get sent to $0$. So the eigenvectors with nonzero eigenvalue are a spanning set of the image. Since eigenvectors with distinct eigenvalues are linearly independent, in this case they also form a basis for the image. – Ben West Dec 13 '16 at 20:04

It's not how many, it's the fact that $B$ has an eigenvalue $0$. Each nonzero vector in the kernel is an eigenvector of $B$ associated to the eigenvalue $0$, so if there's an eigenvalue $0$, the kernel has dimension at least one, and hence the image (by the rank-nullity theorem) has dimension at most two.

Since there are two other distinct nonzero eigenvalues and $B$ is $3\times 3$, its image must have dimension exactly two (and hence its rank is two).

EDIT: Clarification.

Lemma: Let $A$ be an operator and for each $i=1,\dots,n$ let $v_i$ be an eigenvector of $A$ associated to an eigenvalue $\lambda_i$. If the $\lambda_i$ are all distinct,then $\{v_1,\dots,v_n\}$ is linearly independent.

Proof: We do this by induction on $n$. For the base case, we take $n=2$. Let

\begin{align}\tag{1}\label{base1} c_1v_1+c_2v_2=0 \end{align}

be a linear combination of the $v_i$ that equals $0$. Applying $A$ to $\eqref{base1}$ yields

\begin{align}\tag{2}\label{base2} A(c_1v_1+c_2v_2)=\lambda_1c_1v_1+\lambda_2c_2v_2=0 \end{align}

Now, multiply $\eqref{base1}$ by $\lambda_1$ to obtain

\begin{align}\tag{3}\label{base3} \lambda_1c_1v_1+\lambda_1c_2v_2=0 \end{align}

Now, the expression $\eqref{base2}-\eqref{base3}$ becomes $(\lambda_2-\lambda_1)c_2v_2=0$. Since the $\lambda_i$ are all distinct, $(\lambda_2-\lambda_1)\neq0$ and it follows that $c_2=0$. Substituing back into $\eqref{base1}$, we conclude that $c_1$ too is $0$, and hence $\{v_1,v_2\}$ is linearly independent.

Now, suppose the statement is valid for all $m$ with $2\leq m \leq n$. We will show it holds for $n+1$. Indeed, let

\begin{align}\tag{4}\label{induc1} \sum_{i=1}^{n+1}c_iv_i=0 \end{align}

be a linear combination of the $v_i$ that equals $0$. Then

\begin{align}\tag{5}\label{induc2} A\left(\sum_{i=1}^{n+1}c_iv_i\right)=\sum_{i=1}^{n+1}\lambda_ic_iv_i=0 \end{align}

and multiplying $\eqref{induc1}$ by $\lambda_{n+1}$ yields

\begin{align}\tag{6}\label{induc3} \sum_{i=1}^{n+1}\lambda_{n+1}c_iv_i=0 \end{align}

Calculating $\eqref{induc2}-\eqref{induc3}$, the term in $v_{n+1}$ vanishes and we're left with

$$\sum_{i=1}^n (\lambda_i-\lambda_{n+1})c_iv_i=0$$

Here we may apply the induction hypothesis: $\{v_1,\dots,v_n\}$ is linearly indepdent, so the coefficients of the linear combination above must be all $0$. Since the $\lambda_i$ are all distinct, for each $i=1,\dots,n$ we have that $\lambda_i-\lambda_{n+1}\neq0$ and hence it follows that $c_i=0$. Finally, substituting them back into $\eqref{induc1}$ yields $c_{n+1}v_{n+1}=0$, so $c_{n+1}$ too is $0$. It follows that all coefficients of $\eqref{induc1}$ are $0$, and hence $\{v_1,\dots,v_{n+1}\}$ is linearly independent. $\square$

• I don't see the leap from two distinct nonzero eigenvalues to dimension exactly two. – Hello Dec 13 '16 at 19:57
• See the edit. It contains a proof that 'Eigenvectors associated to distinct eigenvalues are linearly independent'. When the operator $A$ is symmetric, they're even orthogonal! It's a nice exercise. – Fimpellizieri Dec 13 '16 at 20:33
• Notice that if $v_i$ is an eigenvector of $B$ associated to an eigenvalue $\lambda_i\neq0$, then $v_i$ is in the image of $B$. Indeed, $$v_i=B\left(\frac{1}{\lambda_i}v_i\right)$$ Hence, if $v_1,\dots,v_n$ are eigenvectors of $B$ associated to distinct and nonzero eigenvalues, then the image of $B$ contains them all and (by the lemma) has dimension at least $n$. Is the connection clearer now? – Fimpellizieri Dec 13 '16 at 20:51
• yes thank you!! – Hello Dec 13 '16 at 21:03

You know that the eigenvector associated with the $0$ eigenvalue is in the kernel of $B$.

And you know that the eigenvectors are independent. The space spanded by the set of eigenvectors associated with the non-zero eigenvalues has dimension equal to the number of non-zero eigenvalues.

• Yes I understand that but what does eigen space of non-zero eigevnalues have anything to do with kernel of B(I CLEARLY understand the 0 part). – Hello Dec 13 '16 at 19:52
• They are not in the kernel! Only vectors parallel to that 0 eigenvector are in the kernel. – Doug M Dec 13 '16 at 19:54
• Most of people who answered are saying two distinct nonzero eigenvalues imply dimension of two. How do you make this leap? – Hello Dec 13 '16 at 20:00
• 2 non-zero eigenvalues indicates 2 independent vectors that are not in the kernel. – Doug M Dec 13 '16 at 20:05

Suppose that a matrix has rank $1$. This means there is a vector $w$ such that $f(u)$ is a multiple of $w$ for all $u\in U$.

Therefore the only eigenvectors in $V$ are the multiples of $U$. So only one non-zero eigenvalue can exist (since if $f(u)=\lambda(u))$ then $f(ku)=\lambda(ku)$.

Since in this case we have more than $1$ non-zero eigenvalue we can conclude the rank is greater than $1$.