## Linear Algebra and Its Applications, Exercise 3.1.11

Exercise 3.1.11. Fredholm’s alternative to the fundamental theorem of linear algebra states that for any matrix $A$ and vector $b$ either

1) $Ax = b$

has a solution or

2) $A^Ty = 0, y^Tb \ne 0$

has a solution, but not both. Show that assuming both (1) and (2) have solutions leads to a contradiction.

Answer: Suppose that both (1) and (2) have solutions for any matrix $A$ and any vector $b$. In other words $Ax = b$ for some $x$ and $A^Ty = 0$ for some $y$, where $y^Tb \ne 0$.

Since $A^Ty = 0$ we also have $(A^Ty)^T = 0$. But $(A^Ty)^T = y^T(A^T)^T = y^TA$ so that we also have $y^TA = 0$. Multiplying both sides by $x$ we have $y^TAx = 0 \cdot x = 0$. But $Ax = b$ so the equation $y^TAx = 0$ reduces to $y^Tb = 0$ contrary to our original assumption that $y^Tb \ne 0$.

We have thus shown that either (1) can have a solution or (2) can have a solution but not both at the same time.

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 .

This entry was posted in linear algebra and tagged , . Bookmark the permalink.