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## How to Find the Standard Matrix of a Linear Transformation?

We have already known that the standard matrix $$A$$ of a linear transformation $$T$$ has the form

$A=[T(\vec{e}_1)\quad T(\vec{e}_2) \quad \cdots \quad T(\vec{e}_n)]$

That means, the $$i$$th column of $$A$$ is the image of the $$i$$th vector of the standard basis. According to this, if we want to find the standard matrix of a linear transformation, we only need to find out the image of the standard basis under the linear transformation.

There are some ways to find out the image of standard basis. Those methods are:

• Find out $$T(\vec{e}_i)$$ directly using the definition of $$T$$;
• Find out $$T(\vec{e}_i)$$ indirectly using the properties of linear transformation, i.e $$T(a \vec{u}+b\vec{v})=aT(\vec{u})+bT(\vec{v})$$;
• Using inverse matrix.

Now we use some examples to illustrate how those methods to be used.

Example 1(find the image directly): Find the standard matrix of linear transformation $$T$$ on $$\mathbb{R}^2$$, where $$T$$ is defined first to rotate each point $$90^\circ$$ and then reflect about the line $$y=x$$.

Solution: First, the rotation $$90^\circ$$ turns $$\vec{e}_1$$ to be $$\vec{e}_2$$ and $$\vec{e}_2$$ to be $$-\vec{e}_1$$. Then reflecting turns $$\vec{e}_2$$ to be $$\vec{e}_1$$ and $$-\vec{e}_1$$ to be $$-\vec{e}_2$$. Hence we have

$T(\vec{e}_1)=\vec{e}_1,\qquad T(\vec{e}_2)=-\vec{e}_2.$

That is

$T(\vec{e}_1)=T\begin{pmatrix}1 \\ 0 \end{pmatrix}= \begin{pmatrix}1\\0\end{pmatrix}, \qquad T(\vec{e}_2)=T\begin{pmatrix}0\\1\end{pmatrix}= \begin{pmatrix}0\\-1\end{pmatrix}$

So the standard matrix is

$A= [T(\vec{e}_1)\quad T(\vec{e}_2)] = \begin{pmatrix}1& 0\\0& -1\end{pmatrix}$

Example 2(find the image using the properties): Suppose the linear transformation $$T$$ is defined as reflecting each point on $$\mathbb{R}^2$$ with the line $$y=2x$$, find the standard matrix of $$T$$.

Solution: Since we can’t find the image of $$\vec{e}_1$$ and $$\vec{e}_2$$ directly, we need some trick to do it. Since any point on the line is unchanged under the transformation, we can choose any point on the line, say $$\begin{pmatrix}1\\2\end{pmatrix}$$, satisfies $$T\begin{pmatrix}1\\2\end{pmatrix}= \begin{pmatrix}1\\2\end{pmatrix}$$ . By the properties of linear transformation, this means

$T\begin{pmatrix}1\\2\end{pmatrix} = T\left( \begin{pmatrix}1\\0\end{pmatrix} +2 \begin{pmatrix}0\\1\end{pmatrix} \right)=T(\vec{e}_1+2\vec{e}_2)= T(\vec{e}_1)+2T(\vec{e}_2)$

So

$T(\vec{e}_1)+2T(\vec{e}_2) = \begin{pmatrix}1\\2\end{pmatrix}$

This is an equation of $$T(\vec{e}_1)$$ and $$T(\vec{e}_2)$$. To solve $$T(\vec{e}_1)$$ and $$T(\vec{e}_2)$$ we need another equation to make the solution unique.

Now we can choose the normal line which passes the origin, $$y=-\frac{1}{2}x$$ and any points on the normal line reflects to the line $$y=2x$$ is equivalent to the reflecting with respect to the origin. So the point $$\begin{pmatrix}-2\\1\end{pmatrix}$$ on the normal line has image $$T \begin{pmatrix}-2\\1\end{pmatrix} =\begin{pmatrix}2\\-1\end{pmatrix}$$, by the properties of linear transformation

$T \begin{pmatrix}-2\\1\end{pmatrix} =-2T(\vec{e}_1)+T(\vec{e}_2)=\begin{pmatrix}2\\-1\end{pmatrix}$

Now we have two equations,

$$T(\vec{e}_1)+2T(\vec{e}_2) = \begin{pmatrix}1\\2\end{pmatrix}$$

$$-2T(\vec{e}_1)+T(\vec{e}_2)=\begin{pmatrix}2\\-1\end{pmatrix}$$

Solve these equations, we have (the first equation minus twice of second equation gives $$T(\vec{e}_1)$$ and the second equation add twice of the first gives $$T(\vec{e}_2)$$ )

$T(\vec{e}_1) =\begin{pmatrix}-\frac{3}{5}\\ \frac{4}{5}\end{pmatrix}, \quad T(\vec{e}_2) = \begin{pmatrix} \frac{4}{5} \\ \frac{3}{5}\\ \end{pmatrix}$

Hence the standard matrix is

$A= \begin{pmatrix}-\frac{3}{5}& \frac{4}{5} \\ \frac{4}{5}& \frac{3}{5} \end{pmatrix}$

The last one related to the method of how to use inverse matrix to find the standard matrix.

Example 3(using inverse matrix to find the standard matrix): Suppose the linear transformation $$T$$ is define by

$T\begin{pmatrix}1\\ 4\end{pmatrix}= \begin{pmatrix}1\\1 \end{pmatrix} \quad T\begin{pmatrix}2\\7\end{pmatrix}= \begin{pmatrix}-1\\1\end{pmatrix},$

find the standard matrix of $$T$$.

Solution: Since for any linear transformation $$T$$ with the standard matrix $$A$$, $$T(\vec{x})=A(\vec{x})$$, we have

$A\begin{pmatrix}1\\ 4\end{pmatrix}= \begin{pmatrix}1\\1 \end{pmatrix} \quad A\begin{pmatrix}2\\7\end{pmatrix}= \begin{pmatrix}-1\\1\end{pmatrix} .$

Equivalently,

$A\begin{pmatrix}1&2\\ 4&7\end{pmatrix}= \begin{pmatrix}1&-1\\1&1 \end{pmatrix} .$

Since

$\begin{vmatrix} 1&2\\ 4&7 \end{vmatrix}=-1\ne 0,$

the matrix

$\begin{pmatrix}1&2\\ 4&7\end{pmatrix}$

is invertible, so we have

$A= \begin{pmatrix}1&-1\\1&1 \end{pmatrix} \begin{pmatrix}1&2\\ 4&7\end{pmatrix}^{-1}.$

Because

$\begin{pmatrix}1&2\\ 4&7\end{pmatrix}^{-1} = \begin{pmatrix}-7&2\\ 4&-1\end{pmatrix},$

we have

$A= \begin{pmatrix}1&-1\\1&1 \end{pmatrix} \begin{pmatrix}-7&2\\ 4&-1\end{pmatrix}= \begin{pmatrix}-11&3\\ -3&1\end{pmatrix} .$

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## How to Determine if a Vector Set is Linearly Dependent or Independent?

By the definition, a vector set is linearly independent if the equality

holds only if all . If one of them is not zero, we say that the vector set is linearly dependent.

If we rewrite the equality as a linear system , where is the matrix that has the vectors as its columns, that is

Then the vector set is linearly independent is equivalent to the system has only trivial solution. And hence we have some equivalent conditions

• are linearly independent
• has only trivial solution
• has unique solution

and some other conditions. From all of those conditions we know that the most efficient way to determine if the vector set is linearly dependent or independent is using the rank. Because using rank, we only need to reduce the matrix to be the row echelon form and no other computation is needed. We don’t need to solve the system, we even don’t need to reduce the matrix to be reduced echelon form. This method need the least computation so we recommend it.

Now let us look an example.

Example: Determine if the vectors are linearly dependent or independent:

Solution: (1) Let

Using row reduction for , we have

The row echelon form has only 2 nonzero rows, Hence . So the vectors are linearly dependent.

(2) Let

The row echelon form has 3 nonzero rows. . So the vectors are linearly independent.