## Linear Algebra and Its Applications, Exercise 2.6.13

Exercise 2.6.13. Suppose that $A$ is a linear transformation from the $x$ $y$ plane to itself. If a transformation $A^{-1}$ exists such that $A^{-1}(Ax) = A(A^{-1}x) = x$ show that $A^{-1}$ is also linear. Also show that if $M$ is the matrix representing $A$ then the matrix representing $A^{-1}$ must be $M^{-1}$.

Answer: Consider the expression $cA^{-1}x + dA^{-1}y$ for any $x$ and $y$. We know from the definition of $A^{-1}$ that $cA^{-1}x + dA^{-1}y = A^{-1}[A(cA^{-1}x + dA^{-1}y)]$

Since $A$ is a linear transformation we then have $A^{-1}[A(cA^{-1}x + dA^{-1}y)] = A^{-1}[cA(A^{-1}x) + dA(A^{-1}y)]$

and using the definition of $A^{-1}$ we have $A(A^{-1}x) = x$ and $A(A^{-1}y) = y$ so that $A^{-1}[cA(A^{-1}x) + dA(A^{-1}y)] = A^{-1}(cx + dy)$

From our final result and initial expression we have $A^{-1}(cx + dy) = cA^{-1}x + dA^{-1}y$

for all $x$ and $y$. So if $A$ is a linear transformation and $A^{-1}$ exists then $A^{-1}$ is a linear transformation also.

Now suppose that $M$ is the matrix representing $A$ and $N$ is the matrix representing $A^{-1}$. By 2V on page 123 the composition of $A$ and $A^{-1}$ is itself a linear transformation, with $AA^{-1}$ represented by the product matrix $MN$ and $A^{-1}A$ represented by the product matrix $NM$.

But since $A^{-1}(Ax) = A(A^{-1}x) = x$ for all $x$ we know that the compositions of $A$ and $A^{-1}$ are equivalent to the identity transformation represented by the identity matrix $I$. We therefore have $MN = I$ and $NM = I$ which implies that $N = M^{-1}$.

So if $M$ is the matrix representing $A$ then $M^{-1}$ is the matrix representing $A^{-1}$.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books .

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### 2 Responses to Linear Algebra and Its Applications, Exercise 2.6.13

1. Stacy says:

hey u free i wanna ask for help with a problem if you wanna.
Nelly’s shadow is 1.5m a tree that is 30 meters tall has a shadow of 25m how tall is Nelly?

• hecker says:

One good place to get answers to general math questions like this is openstudy.com.