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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,

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

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

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 the Type of Discontinuous Points?

A function is continuous at one point \(x=a\) means \(\lim_{x\to a}f(x)=f(a)\). This equation has three meanings: if \(f(x)\) is continuous at point \(x=a\),

  • \(f(a)\) is defined
  • \(\lim_{x\to a}f(x)\) exits
  • \(\lim_{x\to a}f(x)=f(a)\)

So any one of above conditions is not satisfied, the function is discontinue at that point. According to which condition is not satisfied, we can classify the type of discontinuous points.

Since the limit exist at a point is equivalent to the left limit equals right limit, i.e \(\lim_{x\to a^-}f(x)= \lim_{x\to a+}f(x) \), the discontinuity conditions (hence discontinuous points) are classified as

  • \(\lim_{x\to a}f(x)\) exits, \(\lim_{x\to a}f(x)\ne f(a)\) \(\Longrightarrow\) removable discontinuous points;
  • \(\lim_{x\to a}f(x)\) exits, \(f(a)\) is not defined \(\Longrightarrow\) removable discontinuous points;
  • \(\lim_{x\to a^-}f(x) \ne \lim_{x\to a+}f(x) \) but both are finite \(\Longrightarrow\) jump discontinuous points.
  • \(\lim_{x\to a^-}f(x)=\pm\infty\) or \( \lim_{x\to a+}f(x)=\pm\infty\) (or both) \(\Longrightarrow\) infinity discontinuous points;
  • \(\lim_{x\to a^-}f(x)\) or \( \lim_{x\to a+}f(x)\) does not exits nor infinity \(\Longrightarrow\) oscillation discontinuous points.

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What are Common Methods to Evaluate Limits?

There are many methods to evaluate limits. The most common methods are:

  • Limits laws
  • L’Hospital’s principle
  • Some elementary methods, such as factoring, rationalization, compare the order of infinity, etc.

There are many other methods to evaluate limits such as squeeze theorem, the definition of definite integral, Taylor expansion, etc. But for us, those methods above are enough for use.

The limits laws are basic for everyone and we should master it. We don’t want to explain it further. Each time when we evaluate a limit, we should check if the limits laws can be applied. If it is not, we then should use another method to do it.

Now we list some cases and illustrate how to apply those methods.

  • If f(x)=\frac{p(x)}{q(x)} is a rational function, where p(x), q(x) are polynomials and \lim_{x\to a}p(x)=0, \lim_{x\to a}q(x)=0, then they have the common factor x-a, and we can cacel this factor and then evaluate the limit using limit laws. For example,

        \[\lim_{x\to1}\frac{x^2-x+2}{x^2-1}=\lim_{x\to1}\frac{(x+2)(x-1)}{(x+1)(x-1)}=\lim_{x\to1}\frac{x+2}{x+1}=\frac{3}{2}\]

  • If f(x) is a rational function and both p(x), q(x) approaches to infinity, then the method of compare the order of infinity can be applied. For example,

        \[\lim_{x\to\infty}\frac{3x^2+2x-7}{5x^2-4x-1}=\frac{3+2/x-7/x^2}{2-4/x-1/x^2}=\frac{3}{5}.\]

    Here we divided both nominator and denominator by the highest order term x^2. In fact, if the order of the nominator is higher than the the denominator, the result is infinity (positive or negative, depends on the direction and the order). If the order of the nominator is less than the denominator, the result is 0. if the order of the nominator and denominator equals, the result is the ratio of the coefficients of the highest terms.
  • If f(x) has radical terms, it implies rationalization should be applied. For example,

        \begin{align*}\lim_{x\to\infty}(\sqrt{x^2+x}-\sqrt{x^2-x})&=\lim_{x\to\infty}\frac{ (\sqrt{x^2+x}-\sqrt{x^2-x})  (\sqrt{x^2+x}+\sqrt{x^2-x}) }{ \sqrt{x^2+x}+\sqrt{x^2-x} }\\ &=\lim_{x\to\infty}\frac{2x}{ \sqrt{x^2+x}+\sqrt{x^2-x} }\\ &= \lim_{x\to\infty}\frac{2}{ \sqrt{1+1/x}+\sqrt{1-1/x} } =1.\end{align*}

    The last second equality is attained by divide both nominator and denominator by x.
  • L’Hospital’s rule is the widest used method to evaluate indeterminate limits. It can deal with almost all indeterminate limits.
  • For two basic indeterminate forms, \frac{0}{0} and \frac{\infty}{\infty}, this rule can be applied directly.

        \[\lim_{x\to0}\frac{x-\sin x}{x^3}=\lim_{x\to0}\frac{1-\cos x}{3x^2}=\lim_{x\to0}\frac{\sin x}{6x}=\lim_{x\to0}\frac{\cos x}{6}=\frac{1}{6}.\]

  • For the type 0\cdot\infty, we change the form to be \frac{0}{1/\infty} (it becomes the form \frac{0}{0}) or \frac{\infty}{1/0} (it becomes the form \frac{\infty}{\infty}), then applied l’Hospital’s rule.

        \[\lim_{x\to 0^+}x^n\ln x=\lim_{x\to0^+}\frac{\ln x}{\frac{1}{x^n}}=\lim_{x\to0^+}\frac{\frac{1}{x}}{-\frac{n}{x^{n+1}}}=\lim_{x\to0^+}\frac{x^{n+1}}{-nx}=0\]

  • For the type \infty-\infty, we use common denominator or other algebraic operation to transform it to the basic form.

        \[\lim_{x\to\frac{\pi}{2}}(\sec x-\tan x)=\lim_{x\to\frac{\pi}{2}}\frac{1-\sin x}{\cos x}=\lim_{x\to\frac{\pi}{2}}\frac{-\cos x}{-\sin x}=0\]

  • For the types 0^0, \infty^0, 1^{\infty}, we use the identity x=e^{\ln x} to transform the form to be the basic form.

        \[\lim_{x\to0^+}x^x=\lim_{x\to0^+}e^{\ln(x^x)}= \lim_{x\to0^+}e^{x\ln x}=e^{  \lim_{x\to0^+}x\ln x}.  \]

    Since

        \[ \lim_{x\to0^+}x\ln x = \lim_{x\to0^+}\frac{\ln x}{\frac{1}{x}}= \lim_{x\to0^+}\frac{\frac{1}{ x}}{-\frac{1}{x^2}}= \lim_{x\to0^+}-x=0. \]

    So the original limit equals e^0=1.

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

By the definition, a vector set (\vec{u}_1, \vec{u}_2,\cdots, \vec{u}_n) is linearly independent if the equality

    \[k_1 \vec{u}_1+k_2\vec{u}_2+\cdots+k_n\vec{u}_n=0 \]

holds only if all k_i=0, 1\le i\le n. 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 A\vec{x}=\vec{0}, where A is the matrix that has the vectors as its columns, that is

    \[A=\begin{pmatrix}\vec{u}_1&\vec{u}_2&\cdots&\vec{u}_n\end{pmatrix}, \qquad \vec{x}=\begin{pmatrix}k_1\\k_2\\ \vdots\\ k_n\end{pmatrix}.\]

Then the vector set is linearly independent is equivalent to the system A\vec{x}=\vec{0} has only trivial solution. And hence we have some equivalent conditions

  • \vec{u}_1, \vec{u}_2,\cdots, \vec{u}_n are linearly independent
  • A\vec{x}=0 has only trivial solution
  • A\vec{x}=\vec{b} has unique solution
  • \text{det} A=|A|\ne 0
  • \text{Rank}A=n

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:

    \[(1) \begin{pmatrix}-1\\3\\1\end{pmatrix},  \begin{pmatrix}2\\1\\0\end{pmatrix},  \begin{pmatrix}1\\4\\1\end{pmatrix}\qquad (2)  \begin{pmatrix}2\\3\\0\end{pmatrix},  \begin{pmatrix}-1\\4\\0\end{pmatrix},  \begin{pmatrix}0\\0\\2\end{pmatrix} . \]

Solution: (1) Let

    \[ A=\begin{pmatrix}-1&2&1\\3&1&4\\1&0&1\end{pmatrix} \]

Using row reduction for A, we have

    \[ A=\begin{pmatrix}-1&2&1\\3&1&4\\1&0&1\end{pmatrix}\sim   \begin{pmatrix} 1&0&1 \\3&1&4\\  -1&2&1 \end{pmatrix}\sim   \begin{pmatrix} 1&0&1 \\0&1&1\\  0&2&2 \end{pmatrix}\sim  \begin{pmatrix} 1&0&1 \\0&1&1\\  0&0&0 \end{pmatrix}  \]

The row echelon form has only 2 nonzero rows, Hence \text{Rank}A=2<3. So the vectors are linearly dependent.

(2) Let

    \[A= \begin{pmatrix} 2&-1&0 \\3&4&0\\  0&0&2 \end{pmatrix}\sim  \begin{pmatrix} 2&-1&0 \\0&\frac{11}{2}&0\\  0&0&2 \end{pmatrix} \]

The row echelon form has 3 nonzero rows. \text{Rank}A=3. So the vectors are linearly independent.