## Linear Algebra and Its Applications, Exercise 2.4.11

Exercise 2.4.11. Suppose that for a matrix $A$ and any $b$ the system $Ax = b$ always has at least one solution. Show that in this case the system $A^Ty = 0$ has no solution other than $y = 0$.

Answer: Suppose $A$ is an $m$ by $n$ matrix and $b$ is an $m$-element vector. If $x = \begin{bmatrix} x_1&x_2&\cdots&x_n \end{bmatrix}^T$ is a solution for $Ax = b$ then $b$ is a linear combination of the $n$ columns of $A$ with the coefficients being $x_1, x_2, \dotsc, x_m$.

If we can find a solution $x$ for any $b$ in $\mathbf R^m$ then the columns of $A$ span $\mathbf R^m$ since any vector in $\mathbf R^m$ can be expressed as a linear combination of the columns of $A$. The rank of $A$ is therefore $r = m$. (The rank cannot be less than $m$ since then there would not be enough linearly independent columns of $A$ to span $\mathbf R^m$, and cannot be greater than $m$ because it is impossible to have more than $m$ linearly independent vectors in $\mathbf R^m$.)

Now consider the left nullspace $\mathcal N(A^T)$ consisting of all vectors $y$ such that $A^Ty = 0$. The dimension of the left nullspace is $m - r$ where $r$ is the rank of $A$. But from above we have $r = m$ so that the dimension of $\mathcal N(A^T)$ is $m - r = m - m = 0$. The only element of $\mathcal N(A^T)$ is therefore the zero vector, and there are no  solutions to $A^Ty = 0$ other than $y = 0$.

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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