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

Stacy says:

August 16, 2013 at 1:19 pm

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:
  
  September 4, 2013 at 5:15 pm
  
  One good place to get answers to general math questions like this is openstudy.com.