How To Find Inverse Of Linear Transformation

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Inverses of Linear Transformations

$\require{amsmath}$
Observe, that the operation that "does cypher" to a ii-dimensional vector (i.eastward., leaves it unchanged) is also a linear transformation, and plays the role of an identity for $ii \times ii$ matrices under "multiplication". We can verify that this is truthful by speedily examining the ii properties of a linear transformation...

Suppose $I(\bar{ten}) = \bar{ten}$ (i.e., $I$ does not change its input vector.)

Is it true that $I(c\bar{10}) = cI(\bar{x})$ ?

Yep!
$$I(c\bar{x}) = c\bar{x} = cI(\bar{x})$$
Is it truthful that $I(\bar{10}+\bar{y}) = I(\bar{ten}) + I(\bar{y})$ ?

Yes!
$$I(\bar{x}+\bar{y}) = \bar{ten} + \bar{y} = I(\bar{x}) + I(\bar{y})$$

So what does the matrix form of this identity, $I$, look like? Remember, the first and second columns of the matrix form point where the vectors $\begin{pmatrix}1\\0\terminate{pmatrix}$ and $\brainstorm{pmatrix}0\\i\stop{pmatrix}$ get nether the linear transformation.
Since $I$ does zip to its inputs, the aforementioned vectors are the outputs, and hence the columns of the matrix grade. In other words... $$I=\brainstorm{bmatrix}one & 0\\0 & i\end{bmatrix}$$

Now that we accept established the existence of an identity, we tin can talk about inverses.

Given any linear transformation, $M$ (in matrix form), does there exist a linear transformation (which we will call $M^{-ane}$) that volition "disengage" the issue of $Thousand$ ? ...and how exercise nosotros discover it?

First, permit us make certain that if such a $Thou^{-1}$ that will "undo" the outcome of $M$ really exists, then it must be a linear transformation...

Let $c$ be a scalar and $\bar{x}$ exist some arbitrary vector, and start past trying to show $M^{-i}(c\bar{x}) = cM^{-1}(\bar{x})$. Suppose that there is a vector $\bar{z}$ such that
$$Thousand^{-i}(c\bar{x}) = \bar{z}$$
Merely and so (presuming $c \ne 0$),
$$\brainstorm{array}{rcll}
G(\bar{z}) &=& c \bar{10} & \quad \quad \textrm{now divide past $c$}\\\\
\displaystyle{\frac{M({\bar{z}})}{c}} &=& \bar{x} & \quad \quad \textrm{recollect $M$ is linear, so}\\\\
Chiliad \left( \displaystyle{\frac{\bar{z}}{c}} \correct) &=& \bar{x} & \quad \quad \textrm{at present apply $M^{-1}$ to both sides}\\\\
\displaystyle{\frac{\bar{z}}{c}} &=& Thousand^{-one}(\bar{ten}) & \quad \quad \textrm{finally, multiply by $c$}\\\\
\bar{z} &=& cM^{-ane}(\bar{x}) & \quad \quad
\end{array}$$
However, nosotros know from higher up that $K^{-ane}(c\bar{x}) = \bar{z}$, so
$$M^{-one}(c\bar{10}) = cM^{-1}(\bar{x})$$
Fifty-fifty if $c=0$, we can notwithstanding come to this conclusion. Consider the post-obit

Suppose $M^{-1}(0 \cdot \bar{x}) = M^{-1}(\bar{0}) = \bar{z}$ for some vector $\bar{z}$. Then, applying $1000$ we find $Yard(\bar{z}) = \bar{0}$. Annotation, it is petty to demonstrate $Thou(\bar{0}) = \bar{0}$ (consider the matrix production). Further, since $M^{-1}$ exists, at that place must exist a unique input vector for $G$ that produces any given output vector, so $M^{-1}(\bar{0}) = \bar{0}$. Lastly, since we know $Chiliad^{-i}(\bar{x})$ is some vector, so $0 \cdot Thousand^{-1}(\bar{x}) = \bar{0}$. Putting these things together, we have
$$One thousand^{-1}(0 \cdot \bar{x}) = M^{-i}(\bar{0}) = \bar{0} = 0 \cdot 1000^{-one}(\bar{x})$$
Hence, the first property of linear transformations holds for $Grand^{-1}$.

At present, additionally presume that $\bar{y}$ is a second vector. We volition attempt to evidence that $M^{-1}(\bar{x} + \bar{y}) = 1000^{-1}(\bar{ten}) + Thou^{-ane}(\bar{y})$.

Suppose there is a vector $\bar{z}$ such that
$$1000^{-ane}(\bar{ten}) + M^{-i}(\bar{y}) = \bar{z}$$
Simply then, by applying $M$ to both sides, nosotros have
$$\begin{array}{rcll}
M(G^{-ane}(\bar{ten}) + M^{-1}(\bar{y}) &=& M(\bar{z}) & \quad \quad \textrm{now entreatment to the linearity of $M$}\ldots\\\\
M(M^{-1}(\bar{ten})) + M(M^{-1}(\bar{y})) &=& M(\bar{z}) & \quad \quad \textrm{inverses abolish one another, and so}\\\\
\bar{x} + \bar{y} &=& M(\bar{z}) & \quad \quad \textrm{now apply $M^{-1}$ to both sides}\ldots\\\\
M^{-1}(\bar{x} + \bar{y}) &=& G^{-ane}(Yard(\bar{z})) & \quad \quad \textrm{again, inverses abolish one another, then}\\\\
M^{-i}(\bar{x} + \bar{y}) &=& \bar{z} & \quad \quad
\cease{array}$$
However, we know from above that $Yard^{-1}(\bar{10}) + M^{-1}(\bar{y}) = \bar{z}$, so
$$Yard^{-1}(\bar{x} + \bar{y}) = Grand^{-1}(\bar{x}) + M^{-1}(\bar{y})$$
Hence, the second property for $M^{-1}$ to be a linear transformation holds.

With both of the necessary backdrop for being a linear transformation holding for $M^{-ane}$, it must itself be a linear transformation.

Now, that nosotros know what nosotros are after is a linear transformation, let's consider a specific case with regard to how the matrix form for this linear transformation might exist found:

Permit united states of america effort to discover (if it exists) the inverse of $$G=\begin{bmatrix}2 & 3\\5 & seven\end{bmatrix}$$
Being a linear transformation, $M^{-1}$ must then too accept some matrix form -- permit u.s. assume it is given by the following:
$$M^{-one} = \begin{bmatrix}2 & 3\\5 & seven\terminate{bmatrix}^{-1} = \begin{bmatrix}a & b\\c & d\end{bmatrix}$$
Nosotros know that $Grand^{-i}$ volition "undo" $M$, which means for whatever vector $\bar{v}$, we have:
$$(Thousand^{-1} \circ M)(\bar{v}) = M^{-i}(Thou(\bar{v})) = \bar{5}$$
Notice, $(Grand^{-i} \circ Thousand)$ leaves its input vector unchanged -- just that can just mean ane thing... $(K^{-one} \circ M)$ must be the identity matrix!

Thus
$$\begin{bmatrix}a & b\\c & d\finish{bmatrix} \begin{bmatrix}2 & 3\\v & 7\cease{bmatrix} = \begin{bmatrix}one & 0\\0 & i\cease{bmatrix}$$
But then, "multiplying" the matrices to etch the underlying linear transformations, nosotros find
$$\begin{align*}
2a+5b &= 1\\
3a+7b &= 0\\
2c+5d &= 0\\
3c+7d &= 1
\end{align*}$$
We tin can use the starting time two equations to solve for $a$ and $b$. (We solve this organisation in the normal mode, multiplying each equation past well-chosen constants, so that when two equations are added together, 1 of the variables is eliminated.)

$\displaystyle{
\begin{align}
vii \cdot 2a + 7 \cdot 5b &= 7\\
-5 \cdot 3a - 5 \cdot 7b &= 0\\
\end{marshal}}$ $\displaystyle{\begin{marshal}
3 \cdot 2a + 3 \cdot 5b &= 3\\
-2 \cdot 3a - 2 \cdot seven b &= 0\\
\end{align}}$

Calculation the two equations on the left leaves an equation which can be solved for $a$, while calculation the equations on the right leaves an equation that can be solved for $b$.

$$(vii \cdot 2 - five \cdot 3)a = 7 \hspace{0.5in} (3 \cdot five - 2 \cdot vii)b = 3$$

These in turn provide the solutions:

$$a = \frac{7}{7 \cdot 2 - v \cdot 3} \qquad b = \frac{iii}{3 \cdot five - ii \cdot seven}$$
We interruption here for ii reasons. First, note that we left the solution for $a$ and $b$ unsimplified, in that we did not evaluate the denominators. Nosotros will continue to leave these denominators unsimplified throughout the rest of the problem. This is so that nosotros can see the course of our solution as information technology relates to the numbers of our original matrix.

We would like to come up up with a shortcut, if yous volition, to notice the changed of any given matrix quickly. If we simplify things now, spotting that shortcut will be much more than difficult.

Second, notice that everything we have done so far we could have done in the context of congruences $\pmod{n}$ instead of equations (as volition be the case when we start talking about the Hill cypher). Nonetheless, "dividing both sides by some coefficient" in a congruence is handled a chip differently. Instead of dividing (and hence, potentially forming a fraction), we volition multiply both sides by "multiplicative inverse$\pmod{north}$ of the coefficient in question". This will go more clear later on, merely call up that hither lies the disquisitional difference between finding inverses of matrices and finding inverses of matrices$\pmod{n}$.

Using the second 2 equations from our group of iv above, nosotros can similarly solve for $c$ and $d$.

Nosotros beginning multiply by well-chosen constants...

$\displaystyle{\begin{align}
-vii \cdot 2c - vii \cdot 5d &= 0\\
5 \cdot 3c + 5 \cdot 7d &= 5\\
\stop{align}}$ $\displaystyle{\begin{align}
-3 \cdot 2c - 3 \cdot 5d &= 0\\
2 \cdot 3c + 2 \cdot 7d &= 2\\
\end{align}}$

Calculation the equations on the left, we make it at an equation that can be solved for $c$, and adding the equations on the right, we get an equation that tin be solved for $d$.

$$(v \cdot 3 - 7 \cdot 2)c = 5 \hspace{0.5in} (2 \cdot 7 - iii \cdot 5)d = 2$$

Solving these, nosotros meet that
$$c = \frac{5}{5 \cdot 3 - 7 \cdot 2} \qquad d = \frac{2}{2 \cdot vii - 3 \cdot 5}$$
Thus
$$\begin{bmatrix}a & b\\c & d\end{bmatrix} = \begin{bmatrix}\dfrac{7}{7 \cdot 2 - v \cdot 3} & \dfrac{3}{iii \cdot five - 2 \cdot seven}\\
\dfrac{five}{five \cdot three - 7 \cdot 2} & \dfrac{ii}{2 \cdot 7 - 3 \cdot 5}\end{bmatrix}$$
Of course, this looks a little messy, so let's see if we can't tighten it up a bit.

First, notice that if nosotros multiply the upper correct and lower left fractions by $\frac{-1}{-1}$, nosotros can get all of the denominators to look the same.
$$\brainstorm{bmatrix}a & b\\c & d\end{bmatrix} = \begin{bmatrix}\dfrac{7}{ii \cdot 7 - 5 \cdot three} & \dfrac{-3}{2 \cdot vii - 5 \cdot 3}\\
\dfrac{-5}{2 \cdot 7 - v \cdot 3} & \dfrac{2}{two \cdot 7 - 5 \cdot iii}\finish{bmatrix}$$
Now, pulling this common denominator out as a gene sitting outside of the matrix (why tin can we do this?), nosotros have
$$\brainstorm{bmatrix}a & b\\c & d\stop{bmatrix} = \frac{1}{2 \cdot vii - 5 \cdot iii}\begin{bmatrix}7 & -three\\-5 & 2\terminate{bmatrix}$$

As such,
$$\begin{bmatrix}ii & 3\\5 & vii\end{bmatrix}^{-ane} = \dfrac{1}{two \cdot 7 - 5 \cdot three}\begin{bmatrix}7 & -3\\-5 & 2\cease{bmatrix}$$
Since, we never simplified any of the arithmetic, we can now conspicuously see where the numbers went in the final form. This suggests the following "formula" for the matrix class of the changed of a given linear transformation:

If a linear transformation, $1000$, has matrix form
$$M=\begin{bmatrix}ten & y\\z & westward\stop{bmatrix}$$
And then its inverse is given by
$$K^{-ane} = \begin{bmatrix}x & y\\z & w\finish{bmatrix}^{-1} = \dfrac{1}{x \cdot w - z \cdot y}\begin{bmatrix}westward & -y\\-z & x\end{bmatrix}$$
Notice that, depending on the values of $x$, $y$, $z$, and $w$, information technology is possible that nosotros might have a zero in the denominator of the fraction above. This would exist bad in the sense that $K$ would no longer have an inverse. So, the value of the denominator, $(x \cdot w - y \cdot z)$, "determines" whether or not our matrix has an inverse. As such, let united states call this special value the determinant of the matrix and announce it in the following way
$$\textrm{det} = \begin{vmatrix}x & y\\z & westward\terminate{vmatrix} = x \cdot westward - z \cdot y$$
We can then rewrite the inverse of a matrix equally
$$G^{-1} = \brainstorm{bmatrix}ten & y\\z & w\end{bmatrix}^{-1} = \textrm{det}^{-1}\begin{bmatrix}west & -y\\-z & ten\terminate{bmatrix}$$