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Chapter 1: Problem 13

The set of solutions to the system of linear equations $$ \begin{array}{r} x_{1}-2 x_{2}+x_{3}=0 \\ 2 x_{1}-3 x_{2}+x_{3}=0 \end{array} $$ is a subspace of \(R^{3}\). Find a basis for this subspace.

Short Answer

Expert verified
The basis for the subspace of the set of solutions to the given system of linear equations is \( \{ \begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix} \} \).

Step by step solution

01

Write the given system of equations as an augmented matrix

First, we write the given system of linear equations as an augmented matrix, which will help us perform row operations on the system. $$ \left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 2 & -3 & 1 & 0 \end{array}\right] $$
02

Perform row operations to find the row echelon form

To simplify the augmented matrix, we perform the following row operations to achieve a row-echelon form: Operation: \(R_2 - 2R_1 \to R_2\) $$ \left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & 1 & -1 & 0 \end{array}\right] $$
03

Express the remaining variable as a linear combination of the others

From the resulting row-echelon matrix, we have two equations: 1. \(x_1 - 2x_2 + x_3 = 0\) 2. \(x_2 - x_3 = 0\) From the second equation, we can express \(x_3\) as a linear combination of \(x_2\): $$ x_3 = x_2 $$
04

Substitute the expression for x3 into the first equation

Substitute the expression for \(x_3\) found in step 3 into the first equation: $$ x_1 - 2x_2 + x_2 = 0 $$ This simplifies to: $$ x_1 = x_2 $$
05

Express the solution set as a linear combination of basis vectors

Now, express the solution set as a linear combination of basis vectors: $$ \begin{cases} x_1 = x_2 \\ x_2 = x_2 \\ x_3 = x_2 \end{cases} $$ We can rewrite the above system as: $$ \begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix} = x_2\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix} $$
06

Find the basis of the subspace

Since the solution set can be expressed as a linear combination of a single vector, the basis for the subspace is: $$ \{ \begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix} \} $$

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Most popular questions from this chapter

Prove the following properties of the cross product: (a) \(u \times v=-(v \times u)\) (d) \(u \times(v+w)=(u \times v)+(u \times w)\) (b) \(u \times u=0\) for any vector \(u\) (e) \((v+w) \times u=(v \times u)+(w \times u)\) (c) \((k u) \times v=k(u \times v)=u \times(k v)\) \((\mathrm{f}) d(u \times v) \times w=(u \cdot w) v-(v \cdot w) u\)

Let \(W_{1}\) and \(W_{2}\) be subspaces of a vector space \(V\). Prove that \(V\) is the direct sum of \(W_{1}\) and \(W_{2}\) if and only if each vector in \(V\) can be uniquely written as \(x_{1}+x_{2}\), where \(x_{1} \in \mathrm{W}_{1}\) and $x_{2} \in \mathrm{W}_{2}$.

Label the following statements as true or false. (a) If \(V\) is a vector space and \(W\) is a subset of \(V\) that is a vector space, then \(W\) is a subspace of \(V\). (b) The empty set is a subspace of every vector space. (c) If \(V\) is a vector space other than the zero vector space, then \(V\) contains a subspace \(W\) such that \(W \neq V\). (d) The intersection of any two subsets of \(V\) is a subspace of \(V\). (e) An \(n \times n\) diagonal matrix can never have more than \(n\) nonzero entries. (f) The trace of a square matrix is the product of its diagonal entries. (g) Let \(\mathrm{W}\) be the \(x y\)-plane in \(\mathrm{R}^{3}\); that is, $\mathrm{W}=\left\\{\left(a_{1}, a_{2}, 0\right): a_{1}, a_{2} \in R\right\\}\(. Then \)W=R^{2}$.

In each part, determine whether the given vector is in the span of \(S\). (a) \((2,-1,1), \quad S=\\{(1,0,2),(-1,1,1)\\}\) (b) \((-1,2,1), \quad S=\\{(1,0,2),(-1,1,1)\\}\) (c) \((-1,1,1,2), \quad S=\\{(1,0,1,-1),(0,1,1,1)\\}\) (d) \((2,-1,1,-3), \quad S=\\{(1,0,1,-1),(0,1,1,1)\\}\) (e) $-x^{3}+2 x^{2}+3 x+3, \quad S=\left\\{x^{3}+x^{2}+x+1, x^{2}+x+1, x+1\right\\}$ (f) $2 x^{3}-x^{2}+x+3, \quad S=\left\\{x^{3}+x^{2}+x+1, x^{2}+x+1, x+1\right\\}$ (g) $\left(\begin{array}{rr}1 & 2 \\ -3 & 4\end{array}\right), \quad S=\left\\{\left(\begin{array}{rr}1 & 0 \\ -1 & 0\end{array}\right),\left(\begin{array}{ll}0 & 1 \\ 0 & 1\end{array}\right),\left(\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right)\right\\}$ (h) $\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right), \quad S=\left\\{\left(\begin{array}{rr}1 & 0 \\ -1 & 0\end{array}\right),\left(\begin{array}{ll}0 & 1 \\ 0 & 1\end{array}\right),\left(\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right)\right\\}$

Prove: For any complex numbers \(z, w \in \mathbf{C},|z+w| \leq|z|+|w|\)
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