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Construct a nonzero $3 \times 3$ matrix $A$ and a vector $\mathbf{b}$ such that $\mathbf{b}$ is not in $\operatorname{Col} A$

Easiest way to do this is to construct matrix that is in echelon form:$A=\left[\begin{array}{lll}{1} & {1} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {0}\end{array}\right]$

Note that $A$ must have at most 2 pivot columns, because if it had three pivot columns, Col $A$ would be same as $\mathbb{R}^{3}$ and every vector in $\mathbb{R}^{3}$ would be in $\operatorname{Col} A .$

Now it is easy to see that vector$b=\left[\begin{array}{l}{0} \\ {0} \\ {1}\end{array}\right]$ is not a linear combination of columns of $A,$ so it is not in Col $A$

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Gideon I.

Algebra

Chapter 2

Matrix Algebra

Section 8

Subspaces of Rn

Introduction to Matrices

Missouri State University

McMaster University

Lectures

01:32

In mathematics, the absolu…

01:11

02:40

Construct a $3 \times 3$ m…

02:54

Construct a $3 \times 3$ n…

02:27

02:17

With $A$ as in Exercise $3…

04:31

Let $A=\left[\begin{array}…

07:00

Determine the null space o…

01:22

Assume that $A$ and $B$ be…

02:10

02:02

Prove each of the followin…

01:37

So one way we could have a three by three matrix and a vector B Such that B is not in the calm space of a would be as if we constructed, um, a Matrix that in this case has only one pivot column. So an example of this would be 100 and then we'll just have a bunch of zeros. So this is a simple example. Um, but here we see that there is only one pivot column. Um, and we can call this a so that way we know that the column space of a or at least the basis of the column space of a is 100 which means that it could be 200300 But there's always going to be a number here, and then zeros are on the bottom. Um, then let's just pick a B vector to be 010 We know that this is not possible because there's no way that we can make that second value in the second row. There's no way we can make that value one. So this would be a very simple example, but an effective one

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