## Linear Algebra and Its Applications, Exercise 2.3.8

Exercise 2.3.8. Describe the column space of the matrix $A = \begin{bmatrix} 1&2 \\ 3&6 \end{bmatrix}$

and give a basis for it. Do the same for $A^2$.

Answer:The second column of $A$ is twice the first column, so that the two vectors are linearly dependent. The column space consists of any vector of the form $c (1, 3)$ where $c$ is any real number; geometrically the column space is a line passing through the origin and the point $(1, 3)$. The vector $(1, 3)$ serves as a basis for the space.

We have $A^2 = \begin{bmatrix} 1&2 \\ 3&6 \end{bmatrix} \begin{bmatrix} 1&2 \\ 3&6 \end{bmatrix} = \begin{bmatrix} 7&14 \\ 21&42 \end{bmatrix}$

Again we have the second column equal to twice the first, so the two vectors are linearly dependent. Also, we have $(7, 21) = 7(1, 3)$ so that the first column of $A^2$ is a linear combination of the first column of $A$. The matrix $A^2$ therefore has the same column space as $A$ and the vector $(1, 3)$ can serve as its basis.

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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