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Construct a nonzero $3 \times 3$ matrix $A$ and a nonzero vector $\mathbf{b}$ such that $\mathbf{b}$ is in $\mathrm{Col} A,$ but $\mathbf{b}$ is not the same as any one of the columns of $A .$

Construct diagonal matrix $A$$A=\left[\begin{array}{lll}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1}\end{array}\right]$Col $A$ is set of all vectors that are linear combination of columns of $A$$\left[\begin{array}{l}{1} \\ {0} \\ {0}\end{array}\right]+\left[\begin{array}{l}{0} \\ {1} \\ {0}\end{array}\right]+\left[\begin{array}{l}{0} \\ {0} \\ {1}\end{array}\right]=\left[\begin{array}{l}{1} \\ {1} \\ {1}\end{array}\right]$So vector $\left[\begin{array}{l}{1} \\ {1} \\ {1}\end{array}\right]$ is in Col $A$ but is not column of $A$$b=\left[\begin{array}{l}{1} \\ {1} \\ {1}\end{array}\right]$

Algebra

Chapter 2

Matrix Algebra

Section 8

Subspaces of Rn

Introduction to Matrices

Campbell University

Harvey Mudd College

Idaho State University

Lectures

01:32

In mathematics, the absolu…

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01:25

Construct a nonzero $3 \ti…

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Construct a $3 \times 3$ m…

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Let $A=\left[\begin{array}…

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Construct a $3 \times 3$ n…

02:27

01:20

Prove each of the followin…

02:02

01:39

02:17

With $A$ as in Exercise $3…

01:43

in this problem, we want to use a three by three matrix A. So that be is in the column space of a but be is not the same as any of the columns of our Matrix. A. So one way to go about this would be to make a the identity matrix, and we know that the columns of a are linearly independent and the column space of a is the span of the column vectors of a And the span of these three column vectors is actually equal to our three, because any vector in our three can be written as a linear combination of these three vectors thes three column vectors. So now we can choose any B that's in our three as long as it's non zero and also a long as it doesn't match any of the columns of a so we could let be, be equal to the Vector 111 And that linear combination is just the first column vector of a plus. The second column Vector of a plus. The third column vector of a. So this shows that B is in a column space of a and clearly be is not equal to, uh, any of the column vectors of a although it can be. But in this problem, we wanted to choose B, so that s o that it's not. And that completes the problem.

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