Let $A$ and $B$ be $n\times n$ matrices. Then prove that
\[\calN(A)\cap \calN(B) \subset \calN(A+B),\] where $\calN(A)$ is the null space (kernel) of the matrix $A$.

Definition.

Recall that the null space (or kernel) of an $n \times n$ matrix is
\[\calN(A)=\{\mathbf{x}\in \R^n \mid A\mathbf{x}=\mathbf{0}\}.\] The null space $\cal(N)$ is a subspace of the $n$-dimensional vector space $\R^n$.

Proof.

Let $\mathbf{x}$ be an arbitrary vector in the intersection $\calN(A)\cap \calN(B)$.
Then the vector $\mathbf{x}$ belongs to both $\calN(A)$ and $\calN(B)$.
Thus, by definition of the null space, we have
\[A\mathbf{x}=\mathbf{0} \text{ and } B\mathbf{x}=\mathbf{0}.\]

If follows from these equalities that we have
\begin{align*}
(A+B)\mathbf{x}=A\mathbf{x}+B\mathbf{x}=\mathbf{0}+\mathbf{0}=\mathbf{0}.
\end{align*}
Hence $\mathbf{x}$ lies in the null space of the matrix $A+B$, that is, $\mathbf{x}\in \calN(A+B)$.

Since $\mathbf{x}$ is an arbitrary element of $\calN(A)\cap \calN(B)$, we have shown the inclusion
\[\calN(A)\cap \calN(B) \subset \calN(A+B),\] as required.

The post Intersection of Two Null Spaces is Contained in Null Space of Sum of Two Matrices first appeared on Problems in Mathematics.

Let
\[A=\begin{bmatrix}
4 & 1\\
3& 2
\end{bmatrix}\] and consider the following subset $V$ of the 2-dimensional vector space $\R^2$.
\[V=\{\mathbf{x}\in \R^2 \mid A\mathbf{x}=5\mathbf{x}\}.\]

(a) Prove that the subset $V$ is a subspace of $\R^2$.

(b) Find a basis for $V$ and determine the dimension of $V$.

Proof.

(a) $V$ is a subspace of $\R^2$

Note that the equality $A\mathbf{x}=5\mathbf{x}$ can be written as
\[(A-5I)\mathbf{x}=\mathbf{0},\] where $I$ is the $2 \times 2$ matrix.
If we define $B=A-5I$, then the subset $V$ becomes
\begin{align*}
V&=\{\mathbf{x}\in \R^2 \mid B\mathbf{x}=\mathbf{0}\}\\
&=\calN(B),
\end{align*}
the null space of the matrix $B$.
Since any null space is a vector space, this shows that $V$ is a subspace of $\R^2$.

(b) A basis for $V$ and the dimension of $V$

From part (a), we have obtained that
\[V=\calN(B),\] where
\[B=A-5I=\begin{bmatrix}
-1 & 1\\
3& -3
\end{bmatrix}.\] Thus, $V$ is a set of solutions of the system $B\mathbf{x}=\mathbf{0}$.

The augmented matrix for this system is reduced as follows.
\[\left[\begin{array}{rr|r}
-1 & 1 & 0 \\
3 &-3 &0
\end{array} \right] \xrightarrow{R_2+3R_1}
\left[\begin{array}{rr|r}
-1 & 1 & 0 \\
0 & 0 & 0
\end{array} \right] \xrightarrow{-R_1}
\left[\begin{array}{rr|r}
1 & -1 & 0 \\
0 & 0 & 0
\end{array} \right].
\]

From this we see that the general solution $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}$ satisfies
\[x_1=x_2,\] and thus the general solution is
\[\mathbf{x}=\begin{bmatrix}
x_2 \\
x_2
\end{bmatrix}=x_2\begin{bmatrix}
1 \\
1
\end{bmatrix}.\]

Therefore we have
\begin{align*}
V&=\left \{ \mathbf{x} \in \R^2 \quad \middle| \quad \mathbf{x}=x_2\begin{bmatrix}
1 \\
1
\end{bmatrix} \text{ for any } x_2 \in \R \right \}\\
&=\Span\left\{ \begin{bmatrix}
1 \\
1
\end{bmatrix} \right \}.
\end{align*}
Thus the set
\[\left\{ \begin{bmatrix}
1 \\
1
\end{bmatrix} \right \}\] is a spanning set of $V$ and it is linearly independent as it consists of only a single vector.
Therefore
\[\left\{ \begin{bmatrix}
1 \\
1
\end{bmatrix} \right \}\] is a basis for the subspace $V$, and thus the dimension of $V$ is $1$.

(Recall that the dimension of a vector space is the number of vectors in a basis for the vector space.)

The post Prove a Given Subset is a Subspace and Find a Basis and Dimension first appeared on Problems in Mathematics.

Let \[A=\begin{bmatrix}
1 & 1 & 2 \\
2 &2 &4 \\
2 & 3 & 5
\end{bmatrix}.\]

(a) Find a matrix $B$ in reduced row echelon form such that $B$ is row equivalent to the matrix $A$.

(b) Find a basis for the null space of $A$.

(c) Find a basis for the range of $A$ that consists of columns of $A$. For each columns, $A_j$ of $A$ that does not appear in the basis, express $A_j$ as a linear combination of the basis vectors.

(d) Exhibit a basis for the row space of $A$.

Hint.

In part (c), you may use the following theorem.

In part (d), you may use the following theorem.

Solution.

(a) Find a matrix $B$ in reduced roe echelon form such that $B$ is row equivalent to the matrix $A$.

We apply the elementary row operations to the matrix $A$ and obtain
\begin{align*}
A=\begin{bmatrix}
1 & 1 & 2 \\
2 &2 &4 \\
2 & 3 & 5
\end{bmatrix}
\xrightarrow{\substack{R_2-2R_1\\ R_3-2R_1}}
\begin{bmatrix}
1 & 1 & 2 \\
0 &0 &0 \\
0 & 1 & 1
\end{bmatrix}
\xrightarrow{R_2 \leftrightarrow R_3}\\
\begin{bmatrix}
1 & 1 & 2 \\
0 & 1 & 1\\
0 &0 &0
\end{bmatrix}
\xrightarrow{R_1-R_2}
\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 1\\
0 &0 &0
\end{bmatrix}.
\end{align*}
The last matrix is in reduced row echelon form that is row equivalent to $A$.
Thus we set
\[B=\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 1\\
0 &0 &0
\end{bmatrix}.\]

(b) Find a basis for the null space of $A$.

The null space $\calN(A)$ of the matrix is the set of solutions of the homogeneous system $A\mathbf{x}=\mathbf{0}$.
By part (a), the augmented matrix $[A\mid \mathbf{0}]$ is row equivalent to $[B \mid \mathbf{0}]$.
Thus, the solution $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}$ must satisfy
\[x_1=-x_3 \text{ and } x_2=-x_3,\] where $x_3$ is a free variable.
Thus the solutions are given by
\[\mathbf{x}=\begin{bmatrix}
-x_3 \\
-x_3 \\
x_3
\end{bmatrix}=x_3\begin{bmatrix}
-1 \\
-1 \\
1
\end{bmatrix}\] for any number $x_3$.
Hence we have
\begin{align*}
\calN(A)&=\{ \mathbf{x}\in \R^3 \mid \mathbf{x}=x_3\begin{bmatrix}
-1 \\
-1 \\
1
\end{bmatrix} \text{ for any } x_3\in \R \} \\
&=\Span\left\{ \quad\begin{bmatrix}
-1 \\
-1 \\
1
\end{bmatrix}\quad \right \}.
\end{align*}

From this, we deduce that the set
\[\left\{ \quad\begin{bmatrix}
-1 \\
-1 \\
1
\end{bmatrix}\quad \right \}\] is a basis for the null space of $A$.

(c) Find a basis for the range of $A$ that consists of columns of $A$. (Longer version)

Let us write
\[A=\begin{bmatrix}
1 & 1 & 2 \\
2 &2 &4 \\
2 & 3 & 5
\end{bmatrix}=[A_1, A_2, A_3],\] where $A_1, A_2, A_3$ are the column vectors of the matrix $A$.
The range $\calR(A)$ of $A$ is the same as the column space of $A$.
Thus, it suffices to find the maximal number of linearly independent vectors among the column vectors $A_1, A_2, A_3$.
Consider the linear combination
\[x_1A_1+x_2A_2+x_3A_3=\mathbf{0}. \tag{*} \] Then this is equivalent to the homogeneous system
\[A\mathbf{x}=\mathbf{0}\] and we already found the solutions in part (b):
\[x_1=-x_3 \text{ and } x_2=-x_3.\] This tells us the vectors $A_1, A_2, A_3$ are linearly dependent.
For example, $x_1=-1, x_2=-1, x_3=1$ is a nonzero solution of the system (*).
Thus, we have
\[\calR(A)=\Span(A_1, A_2, A_3)=\Span(A_1, A_2).\]

On the other hand, if we consider only $A_1$ and $A_2$, they are linearly independent.
(To see this, you just need to repeat the above argument without $A_3$. This amounts to just ignoring the third columns in the computations.)

Therefore, the set $\{A_1, A_2\}$ is a linearly independent spanning set for the range.
Hence a basis of the range consisting column vectors of $A$ is
\[\{A_1, A_2\}.\]

The only column vector which is not a basis vector is $A_3$.
We already found that $x_1=-1, x_2=-1, x_3=1$ is a nonzero solution of (*).
Thus we have
\[-A_1-A_2+A_3=\mathbf{0}.\] Solving this for $A_3$, we obtain the linear combination for $A_3$ of the basis vectors:
\[A_3=A_1+A_2.\]

(c) A shorter solution using the leading 1 method

Here is a shorter solution which uses the following theorem.

Theorem (leading-1 method): If $B$ is a matrix in reduced row echelon form that is row equivalent to $A$, then all the column vectors of $A$ whose corresponding columns in $B$ have leading 1’s form a basis of the range of $A$.

Looking at the matrix $B$, we see that the first and the second columns of $B$ have leading 1’s. Thus the first and the second column vectors of $A$ form a basis for the range of $A$.

(d) Exhibit a basis for the row space of $A$.

We use the following theorem.

Theorem (Row-space method): If $B$ is a matrix in (reduced) row echelon form that is row equivalent to $A$, then the nonzero rows of $B$ form a basis for the row space of $A$.

We have already found such $B$ in part (a), and the first and the second row vectors are nonzero. Thus they form a basis for the row space of $A$.
Hence a basis for the row space of $A$ is
\[\left\{\,\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}, \begin{bmatrix}
0 \\
1 \\
1
\end{bmatrix} \, \right \}.\]

Comment.

The longer solution of part (c) is essentially the proof of Theorem (leading 1 method).
It is good to know where the theorem came from, but when you solve a problem you may forget and just use the theorem like the shorter solution of (c).

The post Row Equivalent Matrix, Bases for the Null Space, Range, and Row Space of a Matrix first appeared on Problems in Mathematics.

Let $W$ be the subset of $\R^3$ defined by
\[W=\left \{ \mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}\in \R^3 \quad \middle| \quad 5x_1-2x_2+x_3=0 \right \}.\] Exhibit a $1\times 3$ matrix $A$ such that $W=\calN(A)$, the null space of $A$.
Conclude that the subset $W$ is a subspace of $\R^3$.

Solution.

Note that the defining equation $5x_1-2x_2+x_3=0$ can be written as
\[\begin{bmatrix}
5 & -2 & 1 \\
\end{bmatrix}\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}=0.\]

Hence if we put $A=\begin{bmatrix}
5 & -2 & 1 \\
\end{bmatrix}$, then the defining equation becomes
\[A\mathbf{x}=0.\] Hence, we have
\[W=\{x \in \R^3 \mid A\mathbf{x}=0\},\] which is exactly the null space of the $1\times 3$ matrix $A$.
Hence we have proved that $W=\calN(A)$.

In general, the null space of an $m\times n$ matrix is a subspace of the vector space $\R^n$.
(See the post The null space (the kernel) of a matrix is a subspace of $\R^n$.)
Since we showed that $W$ is the null space of $1\times 3$ matrix $A$, we conclude that $W$ is a subspace of $\R^3$.

The post Find a Matrix so that a Given Subset is the Null Space of the Matrix, hence it's a Subspace first appeared on Problems in Mathematics.

Let
\[A=\begin{bmatrix}
1 & 2 & 2 \\
2 &3 &2 \\
-1 & -3 & -4
\end{bmatrix} \text{ and }
B=\begin{bmatrix}
1 & 2 & 2 \\
2 &3 &2 \\
5 & 3 & 3
\end{bmatrix}.\]

Determine the null spaces of matrices $A$ and $B$.

Proof.

The null space of the matrix $A$

We first determine the null space of the matrix $A$.
By definition, the null space is
\[\calN(A):=\{\mathbf{x}\in \R^3 \mid A\mathbf{x}=\mathbf{0}\},\] that is, the null space of $A$ consists of the solution $\mathbf{x}$ of the linear system $A\mathbf{x}=\mathbf{0}$.

To solve the system $A\mathbf{x}=\mathbf{0}$, we apply the Gauss-Jordan elimination. We reduce the augmented matrix $[A|\mathbf{0}]$ by elementary row operations as follows.
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 2 & 2 & 0 \\
2 &3 & 2 & 0 \\
-1 & -3 & -4 & 0
\end{array} \right] \xrightarrow{\substack{R_2-2R_1\\R_3+R_1}}
\left[\begin{array}{rrr|r}
1 & 2 & 2 & 0 \\
0 & -1 & -2 & 0 \\
0 & -1 & -2 & 0
\end{array} \right] \xrightarrow{\substack{R_1+2R_2\\R_3-5R_2}}\\
\left[\begin{array}{rrr|r}
1 & 0 & -2 & 0 \\
0 & -1 & -2 & 0 \\
0 & 0 & 0 & 0
\end{array} \right] \xrightarrow{-R_2}
\left[\begin{array}{rrr|r}
1 & 0 & -2 & 0 \\
0 & 1 & 2 & 0 \\
0 & 0 & 0 & 0
\end{array} \right].
\end{align*}
The last matrix is in reduced row echelon form, and the solutions $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}$ of $A\mathbf{x}=\mathbf{0}$ satisfy
\[x_1=2x_3 \text{ and } x_2=-2x_3\] and $x_3$ is a free variable.
Thus we have
\[\mathbf{x}=\begin{bmatrix}
2x_3 \\
-2x_3 \\
x_3
\end{bmatrix}=x_3\begin{bmatrix}
2 \\
-2 \\
1
\end{bmatrix}\] for any $x_3$ are solutions.

Therefore the null space of the matrix $A$ is
\begin{align*}
\calN(A)&=\left\{\mathbf{x}\in \R^3 \quad \middle| \quad \mathbf{x}=x_3\begin{bmatrix}
2 \\
-2 \\
1
\end{bmatrix} \text{ for any } x_3\in \R \right\}\\
&=\Span\left(\begin{bmatrix}
2 \\
-2 \\
1
\end{bmatrix}\right),
\end{align*}
the subspace of $\R^3$ spanned by the vector $\begin{bmatrix}
2 \\
-2 \\
1
\end{bmatrix}$.

The null space of the matrix $B$

The same procedure works for the matrix $B$. Thus we omit some detail below.
For the matrix $B$, the augmented matrix is reduced to
\[ \left[\begin{array}{rrr|r}
1 & 2 & 2 & 0 \\
2 &3 & 2 & 0 \\
5 & 3 & 3 & 0
\end{array} \right] \to \cdots \to \left[\begin{array}{rrr|r}
1 & 0 & 0 & 0 \\
0 &1 & 0 & 0 \\
0 & 0 & 1 & 0
\end{array} \right].\] (Check the computation by yourself.)
This implies that $\mathbf{x}=\mathbf{0}$ is the only solution of the system $B\mathbf{x}=\mathbf{0}$.
Therefore, the null space $\calN(B)$ of the matrix $B$ consists of just the zero vector:
\[\calN(B)=\left\{\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix} \right\}.\]

Comment.

The dimension of the null space $\calN(A)$ is $1$, and the dimension of the null space $\calN(B)$ is $0$.
In other words, the nullity of the matrix $A$ is $1$, and the nullity of the matrix $B$ is $0$.
(Recall that the nullity of a matrix is just the dimension of the null space of the matrix.)

Related Question.

The proof of the fact that a null space of a matrix is a subspace is give in the post The null space (the kernel) of a matrix is a subspace of $\R^n$.

The post Determine Null Spaces of Two Matrices first appeared on Problems in Mathematics.

Suppose that $n\times n$ matrices $A$ and $B$ are similar.

Then show that the nullity of $A$ is equal to the nullity of $B$.
In other words, the dimension of the null space (kernel) $\calN(A)$ of $A$ is the same as the dimension of the null space $\calN(B)$ of $B$.

Definitions/Hint.

The null space (kernel) of an $m \times n$ matrix $A$ is the subspace of $\R^m$ defined by
\[\calN(A)=\{\mathbf{x} \in \R^m \mid A\mathbf{x}=\mathbf{0}\}.\]

Two matrices $A$ and $B$ are similar if there exists an invertible matrix $S$ such that
\[A=S^{-1}BS.\]

To show that the dimensions of $\calN(A)$ and $\calN(B)$ are equal, find an isomorphism between these vector spaces using the fact that matrices $A$ and $B$ are similar.

Proof.

Since $A$ and $B$ are similar, there exists an invertible matrix $S$ such that
\[A=S^{-1}BS.\] Observe that if $\mathbf{x}\in \calN(A)$, then we have
\begin{align*}
A\mathbf{x}=\mathbf{0}\\
\Leftrightarrow
(S^{-1}BS)\mathbf{x}=\mathbf{0}\\
\Leftrightarrow
B(S\mathbf{x})=\mathbf{0}.
\end{align*}

Therefore we have
\[S\mathbf{x} \in \calN(B).\] From this observation, we define the map
\[\Psi: \calN(A) \to \calN(B)\] by sending $\mathbf{x} \in \calN(A)$ to $S\mathbf{x}\in \calN(B)$.

We claim that the map $\Psi$ is an isomorphism of vector spaces.

To see that $\Psi$ is a linear transformation, let $\mathbf{x}, \mathbf{y} \in \calN(A)$, and $c$ be a scalar.
Then we have
\begin{align*}
\Psi(\mathbf{x}+\mathbf{y})=S(\mathbf{x}+\mathbf{y})=S\mathbf{x}+S\mathbf{y}=\Psi(\mathbf{x})+\Psi(\mathbf{y})
\end{align*}
and
\begin{align*}
\Psi(c\mathbf{x})=S(c\mathbf{x})=cS\mathbf{x}=c\Psi(\mathbf{x}).
\end{align*}
Thus $\Psi$ is a linear transformation.

To show that $\Psi$ is an isomorphism, we give the inverse linear transformation of $\Psi$.

We define $\Phi:\calN(B) \to \calN(A)$ to be a map sending $\mathbf{x} \in \calN(B)$ to $S^{-1}\mathbf{x} \in \calN(A)$.
By a similar argument as above, we can show that the element $S^{-1}\mathbf{x}$ is indeed in $\calN(A)$ and $\Phi$ is a linear transformation and it is straightforward to see that $\Psi\circ \Phi=\id_{\calN(B)}$ and $\Phi \circ \Psi=\id_{\calN(A)}$.
Hence $\Phi$ is the inverse of $\Psi$, and $\Psi$ is an isomorphism.

Therefore the vector spaces $\calN(A)$ and $\calN(B)$ are isomorphic, and hence their dimensions are the same.

Comment.

Instead of finding the inverse linear transformation, you may directly show that the map $\Psi$ is bijective (injective and surjective).

The post Dimension of Null Spaces of Similar Matrices are the Same first appeared on Problems in Mathematics.

In this post, we explain how to diagonalize a matrix if it is diagonalizable.

As an example, we solve the following problem.

Diagonalize the matrix
\[A=\begin{bmatrix}
4 & -3 & -3 \\
3 &-2 &-3 \\
-1 & 1 & 2
\end{bmatrix}\] by finding a nonsingular matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.

(Update 10/15/2017. A new example problem was added.)

Here we explain how to diagonalize a matrix. We only describe the procedure of diagonalization, and no justification will be given.
The process can be summarized as follows. A concrete example is provided below, and several exercise problems are presented at the end of the post.

Diagonalization Procedure

Let $A$ be the $n\times n$ matrix that you want to diagonalize (if possible).

1. Find the characteristic polynomial $p(t)$ of $A$.
2. Find eigenvalues $\lambda$ of the matrix $A$ and their algebraic multiplicities from the characteristic polynomial $p(t)$.
3. For each eigenvalue $\lambda$ of $A$, find a basis of the eigenspace $E_{\lambda}$.
   If there is an eigenvalue $\lambda$ such that the geometric multiplicity of $\lambda$, $\dim(E_{\lambda})$, is less than the algebraic multiplicity of $\lambda$, then the matrix $A$ is not diagonalizable. If not, $A$ is diagonalizable, and proceed to the next step.
4. If we combine all basis vectors for all eigenspaces, we obtained $n$ linearly independent eigenvectors $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n$.
5. Define the nonsingular matrix
   \[S=[\mathbf{v}_1 \mathbf{v}_2 \dots \mathbf{v}_n] .\]
6. Define the diagonal matrix $D$, whose $(i,i)$-entry is the eigenvalue $\lambda$ such that the $i$-th column vector $\mathbf{v}_i$ is in the eigenspace $E_{\lambda}$.
7. Then the matrix $A$ is diagonalized as \[ S^{-1}AS=D.\]

Example of a matrix diagonalization

Now let us examine these steps with an example.
Let us consider the following $3\times 3$ matrix.
\[A=\begin{bmatrix}
4 & -3 & -3 \\
3 &-2 &-3 \\
-1 & 1 & 2
\end{bmatrix}.\] We want to diagonalize the matrix if possible.

Step 1: Find the characteristic polynomial

The characteristic polynomial $p(t)$ of $A$ is
\[p(t)=\det(A-tI)=\begin{vmatrix}
4-t & -3 & -3 \\
3 &-2-t &-3 \\
-1 & 1 & 2-t
\end{vmatrix}.\] Using the cofactor expansion, we get
\[p(t)=-(t-1)^2(t-2).\]

Step 2: Find the eigenvalues

From the characteristic polynomial obtained in Step 1, we see that eigenvalues are
\[\lambda=1 \text{ with algebraic multiplicity } 2\] and
\[\lambda=2 \text{ with algebraic multiplicity } 1.\]

Step 3: Find the eigenspaces

Let us first find the eigenspace $E_1$ corresponding to the eigenvalue $\lambda=1$.
By definition, $E_1$ is the null space of the matrix
\[A-I=\begin{bmatrix}
3 & -3 & -3 \\
3 &-3 &-3 \\
-1 & 1 & 1
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1 & -1 & -1 \\
0 &0 &0 \\
0 & 0 & 0
\end{bmatrix}\] by elementary row operations.
Hence if $(A-I)\mathbf{x}=\mathbf{0}$ for $\mathbf{x}\in \R^3$, we have
\[x_1=x_2+x_3.\] Therefore, we have
\begin{align*}
E_1=\calN(A-I)=\left \{\quad \mathbf{x}\in \R^3 \quad \middle| \quad \mathbf{x}=x_2\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}+x_3\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix} \quad \right \}.
\end{align*}
From this, we see that the set
\[\left\{\quad\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix},\quad \begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}\quad \right\}\] is a basis for the eigenspace $E_1$.
Thus, the dimension of $E_1$, which is the geometric multiplicity of $\lambda=1$, is $2$.

Similarly, we find a basis of the eigenspace $E_2=\calN(A-2I)$ for the eigenvalue $\lambda=2$.
We have
\begin{align*}
A-2I=\begin{bmatrix}
2 & -3 & -3 \\
3 &-4 &-3 \\
-1 & 1 & 0
\end{bmatrix}
\rightarrow \cdots \rightarrow \begin{bmatrix}
1 & 0 & 3 \\
0 &1 &3 \\
0 & 0 & 0
\end{bmatrix}
\end{align*}
by elementary row operations.
Then if $(A-2I)\mathbf{x}=\mathbf{0}$ for $\mathbf{x}\in \R^3$, then we have
\[x_1=-3x_3 \text{ and } x_2=-3x_3.\] Therefore we obtain
\begin{align*}
E_2=\calN(A-2I)=\left \{\quad \mathbf{x}\in \R^3 \quad \middle| \quad \mathbf{x}=x_3\begin{bmatrix}
-3 \\
-3 \\
1
\end{bmatrix} \quad \right \}.
\end{align*}
From this we see that the set
\[\left \{ \quad \begin{bmatrix}
-3 \\
-3 \\
1
\end{bmatrix} \quad \right \}\] is a basis for the eigenspace $E_2$ and the geometric multiplicity is $1$.

Since for both eigenvalues, the geometric multiplicity is equal to the algebraic multiplicity, the matrix $A$ is not defective, and hence diagonalizable.

Step 4: Determine linearly independent eigenvectors

From Step 3, the vectors
\[\mathbf{v}_1=\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}, \mathbf{v}_2=\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}, \mathbf{v}_3=\begin{bmatrix}
-3 \\
-3 \\
1
\end{bmatrix} \] are linearly independent eigenvectors.

Step 5: Define the invertible matrix $S$

Define the matrix $S=[\mathbf{v}_1 \mathbf{v}_2 \mathbf{v}_3]$. Thus we have
\[S=\begin{bmatrix}
1 & 1 & -3 \\
1 &0 &-3 \\
0 & 1 & 1
\end{bmatrix}\] and the matrix $S$ is nonsingular (since the column vectors are linearly independent).

Step 6: Define the diagonal matrix $D$

Define the diagonal matrix
\[D=\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 2
\end{bmatrix}.\] Note that $(1,1)$-entry of $D$ is $1$ because the first column vector $\mathbf{v}_1=\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}$ of $S$ is in the eigenspace $E_1$, that is, $\mathbf{v}_1$ is an eigenvector corresponding to eigenvalue $\lambda=1$.
Similarly, the $(2,2)$-entry of $D$ is $1$ because the second column $\mathbf{v}_2=\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}$ of $S$ is in $E_1$.
The $(3,3)$-entry of $D$ is $2$ because the third column vector $\mathbf{v}_3=\begin{bmatrix}
-3 \\
-3 \\
1
\end{bmatrix}$ of $S$ is in $E_2$.

(The order you arrange the vectors $\mathbf{v}_1, \mathbf{v_2}, \mathbf{v}_3$ to form $S$ does not matter but once you made $S$, then the order of the diagonal entries is determined by $S$, that is, the order of eigenvectors in $S$.)

Step 7: Finish the diagonalization

Finally, we can diagonalize the matrix $A$ as
\[S^{-1}AS=D,\] where
\[S=\begin{bmatrix}
1 & 1 & -3 \\
1 &0 &-3 \\
0 & 1 & 1
\end{bmatrix} \text{ and } D=\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 2
\end{bmatrix}.\] (Here you don’t have to find the inverse matrix $S^{-1}$ unless you are asked to do so.)

Diagonalization Problems and Examples

Check out the following problems about the diagonalization of a matrix to see if you understand the procedure.

For a solution of this problem and related questions, see the post “Diagonalize a 2 by 2 Matrix $A$ and Calculate the Power $A^{100}$“.

For a solution, check out the post “Diagonalize the 3 by 3 Matrix if it is Diagonalizable“.

For a solution, see the post “Quiz 13 (Part 1) Diagonalize a matrix.“.

In the solution given in the post “Diagonalize the 3 by 3 Matrix Whose Entries are All One“, we use an indirect method to find eigenvalues and eigenvectors.

The next problem is a diagonalization problem of a matrix with variables.

The solution is given in the post↴
Diagonalize the Upper Triangular Matrix and Find the Power of the Matrix

A Hermitian Matrix can be diagonalized by a unitary matrix

This means that there exists a unitary matrix $U$ such that $U^{-1}AU$ is a diagonal matrix.

The solution is given in the post ↴
Diagonalize the $2\times 2$ Hermitian Matrix by a Unitary Matrix

More diagonalization problems

More Problems related to the diagonalization of a matrix are gathered in the following page:

Diagonalization of Matrices

The post How to Diagonalize a Matrix. Step by Step Explanation. first appeared on Problems in Mathematics.

Define the map $T:\R^2 \to \R^3$ by $T \left ( \begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}\right )=\begin{bmatrix}
x_1-x_2 \\
x_1+x_2 \\
x_2
\end{bmatrix}$.

(a) Show that $T$ is a linear transformation.

(b) Find a matrix $A$ such that $T(\mathbf{x})=A\mathbf{x}$ for each $\mathbf{x} \in \R^2$.

(c) Describe the null space (kernel) and the range of $T$ and give the rank and the nullity of $T$.

Proof.

(a) Show that $T$ is a linear transformation.

To show that $T: \R^2 \to \R^3$ is a linear transformation, the map $T$ needs to satisfy:

To check (i), let
\[\mathbf{u}=\begin{bmatrix}
u_1 \\
u_2
\end{bmatrix}, \mathbf{v}=\begin{bmatrix}
v_1 \\
v_2
\end{bmatrix}\in \R^2.\] We have
\begin{align*}
T(\mathbf{u}+\mathbf{v})&=T\left( \begin{bmatrix}
u_1+v_1 \\
u_2+v_2
\end{bmatrix} \right) =\begin{bmatrix}
(u_1+v_1)-(u_2+v_2) \\
(u_1+v_1)+(u_2+v_2) \\
u_2+v_2
\end{bmatrix}\\[6pt] &=
\begin{bmatrix}
(u_1-u_2)+(v_1-v_2) \\
(u_1+u_2)+(v_1+v_2) \\
u_2+v_2
\end{bmatrix}
=
\begin{bmatrix}
u_1-u_2\\
u_1+u_2 \\
u_2
\end{bmatrix}+
\begin{bmatrix}
v_1-v_2 \\
v_1+v_2 \\
v_2
\end{bmatrix}\\[6pt] &=T(\mathbf{u})+T(\mathbf{v}).
\end{align*}
Thus condition (i) holds.

To check (ii), consider any $\mathbf{v}=\begin{bmatrix}
v_1 \\
v_2
\end{bmatrix}$ and $c\in \R$.
Then we have
\begin{align*}
T(c\mathbf{v})=&T\left( \begin{bmatrix}
cv_1 \\
cv_2
\end{bmatrix}
\right)
= \begin{bmatrix}
cv_1-cv_2 \\
cv_1+cv_2 \\
cv_2
\end{bmatrix}\\[6pt] &= c\begin{bmatrix}
v_1-v_2 \\
v_1+v_2 \\
v_2
\end{bmatrix}
=cT(\mathbf{v}).
\end{align*}
Thus condition (ii) is met and the map $T$ is a linear transformation from $\R^2$ to $\R^3$.

(b) Find a matrix $A$ such that $T(\mathbf{x})=A\mathbf{x}$ for each $\mathbf{x} \in \R^2$.

(b) Let
\[\mathbf{e}_1=\begin{bmatrix}
1 \\
0
\end{bmatrix}, \mathbf{e}_2=\begin{bmatrix}
0 \\
1
\end{bmatrix}\] be the standard basis vectors of $\R^2$.
The matrix $A$ satisfying $T(\mathbf{x})=A\mathbf{x}$ can be obtained as
\[A=[T(\mathbf{e}_1)\quad T(\mathbf{e}_2)].\] Since we have
\[T(\mathbf{e}_1)=\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix} \text{ and } T(\mathbf{e}_2)=\begin{bmatrix}
-1 \\
1 \\
1
\end{bmatrix},\] we obtain the matrix
\[A=\begin{bmatrix}
1 & -1 \\
1 & 1 \\
0 &1
\end{bmatrix}.\]

(c) Describe the null space (kernel) and the range of $T$ and give the rank and the nullity of $T$

Null Space and Nullity

We fist find the null space of the linear transformation of $T$.
Note that the null space of $T$ is the same as the null space of the matrix $A$.

By definition, the null space is
\[ \calN(T)=\calN(A)=\{\mathbf{x} \in \R^2 \mid A\mathbf{x}=\mathbf{0}\}.\] So the null space is a set of all solutions for the system $A\mathbf{x}=\mathbf{0}$.

To find the solution of the system we consider the augmented matrix and reduce the matrix using the elementary row operations. (The Gauss-Jordan elimination method.)
We have
\begin{align*}
\left[\begin{array}{rr|r}
1 & -1 & 0 \\
1 &1 &0 \\
0 & 1 & 0
\end{array} \right] \xrightarrow{R_2-R_1}
\left[\begin{array}{rr|r}
1 & -1 & 0 \\
0 &2 &0 \\
0 & 1 & 0
\end{array} \right]\\[6pt] \xrightarrow[\text{then} R_2\leftrightarrow R_3] {\substack{R_1+R_3 \\ R_2-2R_1}}
\left[\begin{array}{rr|r}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 0
\end{array} \right].
\end{align*}

Thus the system has only zero solution
\[x_1=0, x_2=0.\] Therefore we obtained
\[\calN(T)=\calN(A)=\left \{ \begin{bmatrix}
0 \\
0
\end{bmatrix} \right \}.\]

Since the nullity is the dimension of the null space, we see that the nullity of $T$ is $0$ since the dimension of the zero vector space is $0$.

Range and Rank

Next, we find the range of $T$. Note that the range of the linear transformation $T$ is the same as the range of the matrix $A$.
We describe the range by giving its basis.

The range of $A$ is the columns space of $A$. Thus it is spanned by columns
\[\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-1 \\
1 \\
1
\end{bmatrix}.\]

From the above reduction of the augmented matrix, we see that these vectors are linearly independent, thus a basis for the range. (Basically, this is the leading 1 method.)
Hence we have
\[\calR(T)=\calR(A)=\Span \left\{\,\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-1 \\
1 \\
1
\end{bmatrix} \,\right\}\] and
\[\left\{\, \begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-1 \\
1 \\
1
\end{bmatrix} \,\right\}\] is a basis for $\calR(T)$.

The rank of $T$ is the dimension of the range $\calR(T)$.
Thus the rank of $T$ is $2$.

Remark that we obtained that the nullity of $T$ is $0$ and the rank of $T$ is $2$. This agrees with the rank-nullity theorem
\[(\text{rank of } T)+ (\text{nullity of } T)=2.\]

More Problems about Linear Transformations

Additional problems about linear transformations are collected on the following page:
Linear Transformation from $\R^n$ to $\R^m$

The post Range, Null Space, Rank, and Nullity of a Linear Transformation from $\R^2$ to $\R^3$ first appeared on Problems in Mathematics.

Let
\[A=\begin{bmatrix}
1 & 1 & 0 \\
1 &1 &0
\end{bmatrix}\] be a matrix.

Find a basis of the null space of the matrix $A$.

(Remark: a null space is also called a kernel.)

Solution.

The null space $\calN(A)$ of the matrix $A$ is by definition
\[\calN(A)=\{ \mathbf{x} \in \R^3 \mid A\mathbf{x}=\mathbf{0} \}.\] In other words, the null space consists of all solutions $\mathbf{x}$ of the matrix equation $A\mathbf{x}=\mathbf{0}$.

So we first determine the solutions of $A\mathbf{x}=\mathbf{0}$ by Gauss-Jordan elimination. The augmented matrix is
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 1 & 0 & 0 \\
1 & 1 & 0 & 0
\end{array} \right].
\end{align*}
Subtracting $R_1$ from $R_2$, we reduce the augmented matrix to the reduced row echelon form matrix as follows.
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 1 & 0 & 0 \\
1 & 1 & 0 & 0
\end{array} \right] \xrightarrow{R_2-R_1}
\left[\begin{array}{rrr|r}
1 & 1 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array} \right].
\end{align*}
Thus the solution $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}$ of $A\mathbf{x}=\mathbf{0}$ satisfies $x_1+x_2=0$, or equivalently $x_1=-x_2$.
Substituting the last equality, we see that solutions are of the form
\[\mathbf{x}=\begin{bmatrix}
-x_2 \\
x_2 \\
x_3
\end{bmatrix}=x_2\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}+x_3\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}.\] Therefore the null space is
\begin{align*}
\calN(A)&=\left \{\mathbf{x} \in \R^3 \quad \middle| \quad \mathbf{x}=x_2\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}+x_3\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix} \text{ for any } x_2, x_3 \in \R \right \}\\[6pt] &= \mathrm{Sp} \left\{ \, \begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}, \,
\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix} \, \right\}.
\end{align*}
Thus, the set $\left\{ \, \begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}, \,
\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix} \, \right\}$ is a spanning set for the null space $\calN(A)$.

Now, we check that the vectors $\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix},
\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}$ are linearly independent.
Consider a linear combination
\[a_1\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}+a_2\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix} =\mathbf{0}.\] This is equivalent to
\[\begin{bmatrix}
-a_1 \\
a_1 \\
a_2
\end{bmatrix}=\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}.\] Hence we must have $a_1=a_2=0$.
Since the linear combination equation has only the zero solution, the vectors $\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix},
\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}$ are linearly independent.

Therefore the set $\left\{ \, \begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}, \,
\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}\, \right\}$ is a linearly independent spanning set, thus it is a basis for the null space $\calN(A)$.

The post Find a Basis For the Null Space of a Given \times 3$ Matrix first appeared on Problems in Mathematics.

Let $A$ be an $m \times n$ real matrix. Then the null space $\calN(A)$ of $A$ is defined by
\[ \calN(A)=\{ \mathbf{x}\in \R^n \mid A\mathbf{x}=\mathbf{0}_m\}.\] That is, the null space is the set of solutions to the homogeneous system $A\mathbf{x}=\mathbf{0}_m$.

Prove that the null space $\calN(A)$ is a subspace of the vector space $\R^n$.
(Note that the null space is also called the kernel of $A$.)
 

Proof.

Subspace criteria

We check the following criteria for $\calN(A)$ to be a subspace of $\R^n$.

Check (a)

Since we have $A\mathbf{0}_n=\mathbf{0}_m$, the zero vector $\mathbf{0}_n \in \R^n$ is in $\calN(A)$. Thus (a) is met.

Check (b)

For (b), take arbitrary vectors $\mathbf{x}, \mathbf{y} \in \calN(A)$.
By the definition of $\calN(A)$, we have
\[A\mathbf{x}=\mathbf{0}_m \text{ and } A\mathbf{y}=\mathbf{0}_m. \tag{*}\] We want to show that $\mathbf{x}+ \mathbf{y} \in \calN(A)$. Thus we need to show $A(\mathbf{x}+\mathbf{y})=\mathbf{0}$.

This equality can be proved as follows.
We have
\begin{align*}
A(\mathbf{x}+\mathbf{y}) &= A\mathbf{x}+A\mathbf{y}\\
&=\mathbf{0}+\mathbf{0}=\mathbf{0}.
\end{align*}
Here the second equality follows from (*).
Thus $\mathbf{x}+\mathbf{y}\in \calN(A)$ and (b) is satisfied.

Check (c)

To check (c), take arbitrary $\mathbf{x} \in \calN(A)$ and a scalar $c \in \R$.
Since $\mathbf{x} \in \calN(A)$, we have
\[A\mathbf{x}=\mathbf{0}_m.\]

Multiplying by the scalar $c$, we have
\[cA\mathbf{x}=c\mathbf{0}_m.\] This is equivalent to
\[A(c\mathbf{x})=\mathbf{0}_m.\] This equality implies that $c\mathbf{x} \in \calN(A)$, hence (c) is met.

Thus we checked the conditions (a), (b), and (c) and conclude that the null space $\calN(A)$ is a subspace of $\R^n$.

Related Question.

In the following problem, we determine the null space of a matrix explicitly.

Determine null spaces of two matrices

The post The Null Space (the Kernel) of a Matrix is a Subspace of $\R^n$ first appeared on Problems in Mathematics.