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Show that if the columns of Bare linearly dependent, then so are the columns of AB.

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The columns of ABare linearly dependent.

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01

The columns of B are linearly dependent

When the columns of Bare linearly dependent, there exists a nonzero vector \(x\) such that \(Bx = 0\).

02

Show the columns of AB are linearly dependent

Therefore,

\(\begin{aligned}{l}A\left( {Bx} \right) = A \times 0\\\left( {AB} \right)x = 0\,\left( {By\,Associative\,law} \right)\end{aligned}\)

The columns of ABmust be linearly dependent since \(x\) is nonzero.

Thus, the columns of ABare linearly dependent.

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Most popular questions from this chapter

Suppose Tand U are linear transformations from \({\mathbb{R}^n}\) to \({\mathbb{R}^n}\) such that \(T\left( {U{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\) . Is it true that \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\)? Why or why not?

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