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.

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