For $n\times n$ square matrix $A$, we have several conditions for the existence of $A^{-1}$. Then the following statements are equivalent:

1.	$A$ is non-singular.
2.	$A{\mathbf x}={\mathbf 0}$ has only the trivial solution ${\mathbf x}={\mathbf 0}$.
3.	If $U$ is a reduced echelon form for $A$, then $U$ has no row of all zeros.
4.	$A{\mathbf x}={\mathbf b}$ has a solution for every $n$-dimensional column vector ${\mathbf b}$.

Then the following statements are equivalent:

1.	$A$ is non-singular.
2.	$A{\mathbf x}={\mathbf 0}$ has only the trivial solution ${\mathbf x}={\mathbf 0}$.
3.	If $U$ is a reduced echelon form for $A$, then $U$ has no row of all zeros.
4.	$A{\mathbf x}={\mathbf b}$ has a solution for every $n$-dimensional column vector ${\mathbf b}$.
5.	det$(A)\ne 0$.
6.	The columns of $A$ are linearly independent.
7.	The rows of $A$ are linearly independent.
8.	nullity($A$) = 0.
9.	rank($A$) = $n$.

Let $A$ be a square matrix. Then an eigenvector of $A$ is a vector ${\mathbf v}\ne {\mathbf 0}$ such that $$ A{\mathbf v} = \lambda {\mathbf v}, $$ for some scalar $\lambda$. We call $\lambda$ the eigenvalue corresponding to ${\mathbf v}$. If ${\mathbf v}$ is an eigenvector of $A$, then so is $t{ \v}$ for any scalar $t\ne 0$.

We have $$ A{\mathbf v}=\lambda{\mathbf v}=\lambda I{\mathbf v}\ \ \Rightarrow \ \ (A-\lambda I){\mathbf v}={\mathbf 0}. $$

We have $$ A{\mathbf v}=\lambda{\mathbf v}=\lambda I{\mathbf v}\ \ \Rightarrow \ \ (A-\lambda I){\mathbf v}={\mathbf 0}. $$

Hence ${\mathbf x}={\mathbf v}$ is a non-trivial solution to the homogeneous system of equations $(A-\lambda I){\mathbf x}={\mathbf 0}$, and conversely, if there is a non-trivial solution then $\lambda$ is an eigenvalue of $A$.

For an $n\times n$ matrix $A$, det$(A-\lambda I)$ is a polynomial of degree $n$ in $\lambda$, called the characteristic polynomial of $A$. The equation det$(A-\lambda I)=0$ is the characteristic equation of $A$.

Note that eigenvalues $\lambda$ may be complex numbers, and the eigenvectors ${\mathbf v}$ may have complex components, even for real matrices $A$.

Note that eigenvalues $\lambda$ may be complex numbers, and the eigenvectors ${\mathbf v}$ may have complex components, even for real matrices $A$.

For example, consider \[ W = \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right) \]

The charateristic equation is

To find the eigenvalues and eigenvectors, do the following:

1. Find the roots of the characteristic polynomial, $\det(A-\lambda I)=0$. These are the eigenvalues.
2. For each eigenvalue $\lambda$, find $N(A-\lambda I),$ known as the eigenspace associated to $\lambda$.

Find the eigenvalues and eigenvectors of $A= \left(\begin{array}{rrr}-3&1&0\\1&-2&1\\0&1&-3\end{array}\right).$

Find the eigenvalues and eigenvectors of $A= \left(\begin{array}{rrr}-3&1&0\\1&-2&1\\0&1&-3\end{array}\right).$

Here we have that

Thus the eigenvalues are $\lambda_1 = -1, \lambda_2=-3,$ and $\lambda_3=-4.$

For $\lambda_1 = -1:$ $\ds \left(A-\lambda_1 I\right) \v=\mathbf 0 $ $ \;\Ra \; \left( \begin{array}{ccc} -2 & 1 & 0 \\ 1 & -1 & 1 \\ 0 & 1 & -2 \\ \end{array} \right) \left( \begin{array}{c} v_1 \\ v_2 \\ v_3 \\ \end{array} \right) =\left( \begin{array}{c} 0 \\ 0 \\ 0 \\ \end{array} \right) $

Substitute into (1) and (2) $\; \Ra \; \left\{ \begin{array}{ccccc} -2v_1 & + & 2v_3 & = & 0 \\ v_1 & - & v_3 & = & 0 \\ \end{array} \right. . $

Substitute into (1) and (2) $\; \Ra \; \left\{ \begin{array}{ccccc} -2v_1 & + & 2v_3 & = & 0 \\ v_1 & - & v_3 & = & 0 \\ \end{array} \right. . $  
Then $v_1 = v_3,$

For $\lambda_2 = -3:$ $\ds \left(A-\lambda_2 I\right) \v=\mathbf 0 $ $ \;\Ra \; \left( \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0 \\ \end{array} \right) \left( \begin{array}{c} v_1 \\ v_2 \\ v_3 \\ \end{array} \right) =\left( \begin{array}{c} 0 \\ 0 \\ 0 \\ \end{array} \right) $

Thus $ \;\left( \begin{array}{c} v_1 \\ v_2 \\ v_3 \\ \end{array} \right) = \left( \begin{array}{c} v_1 \\ 0 \\ -v_1 \\ \end{array} \right) $ $ = v_1 \left( \begin{array}{r} 1 \\ 0 \\ -1 \\ \end{array} \right). $ The eigenvector is $ \left( \begin{array}{r} 1 \\ 0 \\ -1 \\ \end{array} \right). $

For $\lambda_3 = -4:$ $\ds \left(A-\lambda_3 I\right) \v=\mathbf 0 $ $ \;\Ra \; \left( \begin{array}{ccc} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 1 \\ \end{array} \right) \left( \begin{array}{c} v_1 \\ v_2 \\ v_3 \\ \end{array} \right) =\left( \begin{array}{c} 0 \\ 0 \\ 0 \\ \end{array} \right) $

Substitute into (2) $ \;\Ra v_2 + v_3 = 0, $ which is also equation (3).

Substitute into (2) $ \;\Ra v_2 + v_3 = 0, $ which is also equation (3).
Then $v_3 = -v_2,$

Find the eigenvalues and eigenvectors of $A= \left(\begin{array}{rrr}-3&1&0\\1&-2&1\\0&1&-3\end{array}\right).$

For a square matrix $A$:

1. $A$ and $A^T$ have the same eigenvalues.

$\text{det}\left(A- \lambda I\right)$ $= \text{det}\left(\left(A- \lambda I\right)^T\right)$

$\qquad \qquad = \text{det}\left(A^T- \lambda I^T\right)$ $= \text{det}\left(A^T- \lambda I\right).$

2. $A$ is singular if and only if $\lambda=0$ is an eigenvalue of $A.$

$A$ is singular if and only if $A\mathbf x = \mathbf 0$ has a non-trivial solution $\iff \lambda =0 $ is an eigenvalue of $A.$

3. If $\lambda$ is an eigenvalue of $A,$ then $\lambda^2$ is an eigenvalue of $A^2,$ and $1/{\lambda}$ is an eigenvalue of $A^{-1}$ when $A$ is non-singular.

Exercise: 📝

4. If $\lambda$ is an eigenvalue of $A,$ then $\lambda-m$ is an eigenvalue of $A-mI,$ for any scalar $m.$

Exercise: 📝

If $\lambda_1, \lambda_2,\ldots,\lambda_k$ are distinct eigenvalues of $A$, with corresponding eigenvectors $\v_1, \v_2,\ldots,\v_k$ (such that $\v_i$ corresponds to $\lambda_i$), then the set of eigenvectors $\{ \v_1,\v_2,\ldots,\v_k \}$ is linearly independent.

Proof: For $k=1,$ the set $\{\v_1\}$ is linearly independent. Easy! 😃

Assume that $\left\{\v_1, \v_2, \ldots, \v_n\right\}$ is linearly independent. This is our induction hypothesis! Easy as well! 😃

Now we have to prove the case for $n+1$. Set $$a_1\v_1+ a_2\v_2+ \cdots + a_{n+1}\v_{n+1}= \mathbf 0 \qquad (\star)$$

Proof:

Now we have to prove the case for $n+1$. Set $$a_1\v_1+ a_2\v_2+ \cdots + a_{n+1}\v_{n+1}= \mathbf 0 \qquad (\star)$$

Multiply ($\star$) by the matrix $A$ from the left to obtain \[ \begin{multline*} A\left(a_1\v_1+ \cdots + a_{n+1}\v_{n+1}\right) \\ = a_1\lambda_1\v_1+ \cdots + a_{n}\lambda_{n}\v_{n} + a_{n+1}\lambda_{n+1}\v_{n+1} = \mathbf 0\quad (1) \end{multline*} \] Now multiply ($\star$) by $\lambda_{n+1}\neq 0$ to get \[ a_1\lambda_{n+1}\v_1+ a_2\lambda_{n+1}\v_2+ \cdots + a_{n}\lambda_{n+1}\v_{n} + a_{n+1}\lambda_{n+1}\v_{n+1} = \mathbf 0\qquad (2) \]

Proof:

\[ a_1\lambda_1\v_1+ \cdots + a_{n}\lambda_{n}\v_{n} + a_{n+1}\lambda_{n+1}\v_{n+1} = \mathbf 0\qquad (1) \] \[ a_1\lambda_{n+1}\v_1+ a_2\lambda_{n+1}\v_2+ \cdots + a_{n}\lambda_{n+1}\v_{n} + a_{n+1}\lambda_{n+1}\v_{n+1} = \mathbf 0\qquad (2) \]

Substracting (2) - (1) we obtain \[ a_1\left(\lambda_{n+1} -\lambda_1\right)\v_1+ a_2\left(\lambda_{n+1} -\lambda_2\right)\v_2+\cdots + a_n\left(\lambda_{n+1} -\lambda_n\right)\v_n=\mathbf 0. \]

Proof:

Substracting (2) - (1) we obtain \[ a_1\left(\lambda_{n+1} -\lambda_1\right)\v_1+ a_2\left(\lambda_{n+1} -\lambda_2\right)\v_2+\cdots + a_n\left(\lambda_{n+1} -\lambda_n\right)\v_n=\mathbf 0. \]

Recall that $\left\{\v_1, \v_2, \ldots, \v_n\right\}$ is linearly independent by induction hypotesis.

Then $a_1\left(\lambda_{n+1} -\lambda_1\right) = a_2\left(\lambda_{n+1} -\lambda_2\right) = \cdots = a_n\left(\lambda_{n+1} -\lambda_n\right)=0.$ Since $\{\lambda_i\}$ are distinct, then $$a_1=a_2=\cdots = a_n=0.$$

Proof:

So we have that $a_i=0$ for $i=1,\ldots,n$. Substituting in $$a_1\v_1+ a_2\v_2+ \cdots + a_{n+1}\v_{n+1}= \mathbf 0,\quad (\star)$$ we have that $a_{n+1}\v_{n+1}= \mathbf 0\;$ $\;\Ra\; a_{n+1} = 0.$

Thus $\left\{\v_1, \v_2, \ldots, \v_{n+1}\right\}$ is linearly independent. Hence, the set of eigenvectors $$\left\{\v_1, \v_2, \ldots, \v_{k}\right\}$$ is linearly independent. $\quad\blacksquare$