# Linear Transformation, Basis For the Range, Rank, and Nullity, Not Injective

## Problem 276

Let $V$ be the vector space of all $2\times 2$ real matrices and let $P_3$ be the vector space of all polynomials of degree $3$ or less with real coefficients.

Let $T: P_3 \to V$ be the linear transformation defined by

\[T(a_0+a_1x+a_2x^2+a_3x^3)=\begin{bmatrix}

a_0+a_2 & -a_0+a_3\\

a_1-a_2 & -a_1-a_3

\end{bmatrix}\]
for any polynomial $a_0+a_1x+a_2x^2+a_3 \in P_3$.

Find a basis for the range of $T$, $\calR(T)$, and determine the rank of $T$, $\rk(T)$, and the nullity of $T$, $\nullity(T)$.

Also, prove that $T$ is not injective.

Add to solve later

Sponsored Links

## Proof.

Let $\{1, x, x^2, x^3\}$ be a basis of the vector space $P_3$.

Then the range of $T$ is spanned by the images of these basis vectors. That is,

\[\calR(T)=\Span \{T(1), T(x), T(x^2), T(x^3)\}.\]

Thus, the range $\calR(T)$ is spanned by the following matrices

\begin{align*}

T(1)=\begin{bmatrix}

1 & -1\\

0& 0

\end{bmatrix}, T(x)=\begin{bmatrix}

0 & 0\\

1& -1

\end{bmatrix}\\[6pt]
T(x^2)=\begin{bmatrix}

1 & 0\\

-1& 0

\end{bmatrix},

T(x^3)=\begin{bmatrix}

0 & 1\\

0& -1

\end{bmatrix}.

\end{align*}

We find a basis among these matrices.

Let $B=\{E_{11}, E_{12}, E_{21}, E_{22} \}$ be the standard basis of $V$, where the $(i,j)$-entry of $E_{ij}$ is $1$ and the other entries are $0$.

With respect to this basis $B$, the coordinate vectors of the above matrices are

\begin{align*}

[T(1)]_B=\begin{bmatrix}

1 \\

-1 \\

0 \\

0

\end{bmatrix},

[T(x)]_B=\begin{bmatrix}

0 \\

0 \\

1 \\

-1

\end{bmatrix}\\[6pt] [T(x^2)]_B=\begin{bmatrix}

1 \\

0 \\

-1 \\

0

\end{bmatrix},

[T(x^3)]_B=\begin{bmatrix}

0 \\

1 \\

0 \\

-1

\end{bmatrix}.

\end{align*}

Let $A$ be the matrix whose columns are these vectors:

\[A=\begin{bmatrix}

1 & 0 & 1 & 0 \\

-1 &0 & 0 & 1 \\

0 & 1 & -1 & 0 \\

0 & -1 & 0 & -1

\end{bmatrix}.\] Applying the elementary row operations, we obtain

\begin{align*}

A=\begin{bmatrix}

1 & 0 & 1 & 0 \\

-1 &0 & 0 & 1 \\

0 & 1 & -1 & 0 \\

0 & -1 & 0 & -1

\end{bmatrix}

\xrightarrow{\substack{R_2+R_1 \\ R_4+R_3}}

\begin{bmatrix}

1 & 0 & 1 & 0 \\

0 &0 & 1 & 1 \\

0 & 1 & -1 & 0 \\

0 & 0 & -1 & -1

\end{bmatrix}\\[10pt] \xrightarrow{\substack{R_2 \leftrightarrow R_3 }}

\begin{bmatrix}

1 & 0 & 1 & 0 \\

0 & 1 & -1 & 0 \\

0 &0 & 1 & 1 \\

0 & 0 & -1 & -1

\end{bmatrix}

\xrightarrow{\substack{R_1-R_3\\ R_2+R_3\\ R_4+R_3}}

\begin{bmatrix}

1 & 0 & 0 & -1 \\

0 & 1 & 0 & 1 \\

0 &0 & 1 & 1 \\

0 & 0 & 0 & 0

\end{bmatrix}.

\end{align*}

The last matrix is in reduced row echelon form and the first three columns contains the leading $1$’s.

Thus, the first three matrices $T(1), T(x), T(x^2)$ form a basis for the range $\calR(T)$.

(This is the “leading 1 method”.)

(Note also that the last column gives the coefficients of the linear combination $T(x^3)=-T(1)+T(x)+T(x^2)$.)

It follows that the rank, which is the dimension of the range, is $3$.

By the rank-nullity theorem we have

\[\rk(T)+\nullity(T)=\dim(P_3)=4.\]
We deduce that the nullity is $1$.

Finally, since the nullity of $T$ is $1$, the linear transformation $T$ is not injective.

(Recall that $T$ is injective if and only if the nullity is $0$.)

Another way to see that $T$ is not injective is as follows.

We saw that $T(x^3)=-T(1)+T(x)+T(x^2)$. This implies by linearity that

\[T(x^3)=T(-1+x+x^2).\]
Thus the values of $T$ at two distinct polynomials $x^3$ and $-1+x+x^2$ are the same.

Hence $T$ is not injective.

In conclusion,

\[\{T(1), T(x), T(x^2)\}=\left \{\begin{bmatrix}

1 & -1\\

0& 0

\end{bmatrix}, \begin{bmatrix}

0 & 0\\

1& -1

\end{bmatrix},

\begin{bmatrix}

1 & 0\\

-1& 0

\end{bmatrix}, \right \}

\] is a basis for the range of $T$. We have

\begin{align*}

\rk(T)=3 \text{ and } \nullity(T)=1.

\end{align*}

Add to solve later

Sponsored Links