Find the span of ${v}_{1},{v}_{2},...,{v}_{k},w$ when ${x}_{1}\cdot {v}_{1}+{x}_{2}\cdot {v}_{2}+...+{x}_{k}\cdot {v}_{k}=w$ has no solution

kjukks1234531
2022-09-26
Answered

Find the span of ${v}_{1},{v}_{2},...,{v}_{k},w$ when ${x}_{1}\cdot {v}_{1}+{x}_{2}\cdot {v}_{2}+...+{x}_{k}\cdot {v}_{k}=w$ has no solution

You can still ask an expert for help

Ashly Sanford

Answered 2022-09-27
Author has **9** answers

Notice that

$\sum _{m=1}^{k}{x}_{m}{\mathbf{v}}_{m}=\mathbf{u}$

has only one solution. This would imply the set of vectors $\{{\mathbf{v}}_{1},\dots ,{\mathbf{v}}_{k}\}$ are linearly independent. Otherwise one can write one vector vi as linear combination of other vectors and there will be more than one solution (you may try to write out the other solutions explicitly to verify the above argument).

$\sum _{m=1}^{k}{x}_{m}{\mathbf{v}}_{m}=\mathbf{u}$

has only one solution. This would imply the set of vectors $\{{\mathbf{v}}_{1},\dots ,{\mathbf{v}}_{k}\}$ are linearly independent. Otherwise one can write one vector vi as linear combination of other vectors and there will be more than one solution (you may try to write out the other solutions explicitly to verify the above argument).

Haven Kerr

Answered 2022-09-28
Author has **1** answers

Hint: Try to prove that if the vectors $\{{\mathbf{v}}_{1},\dots ,{\mathbf{v}}_{k}\}$ are linearly dependent then

$\sum _{m=1}^{k}{x}_{m}{\mathbf{v}}_{m}=\mathbf{u}$

has either no solution or more than one solutions for every $\mathbf{u}$ in $\mathbb{V}$. The above is just the contrapositive argument.

$\sum _{m=1}^{k}{x}_{m}{\mathbf{v}}_{m}=\mathbf{u}$

has either no solution or more than one solutions for every $\mathbf{u}$ in $\mathbb{V}$. The above is just the contrapositive argument.

asked 2021-02-11

Let F be a fixed 3x2 matrix, and let H be the set of all matrices A in $M}_{2\times 4$ with the property that FA = 0 (the zero matrix in ${M}_{3\times 4})$ . Determine if H is a subspace of $M}_{2\times 4$

asked 2021-05-29

Which of the following expressions are meaningful? Which are meaningless? Explain.

a)$(a\cdot b)\cdot c$

$(a\cdot b)\cdot c$ has ? because it is the dot product of ?.

b)$(a\cdot b)c$

$(a\cdot b)c$ has ? because it is a scalar multiple of ?.

c)$|a|(b\cdot c)$

$|a|(b\cdot c)$ has ? because it is the product of ?.

d)$a\cdot (b+c)$

$a\cdot (b+c)$ has ? because it is the dot product of ?.

e)$a\cdot b+c$

$a\cdot b+c$ has ? because it is the sum of ?.

f)$|a|\cdot (b+c)$

$|a|\cdot (b+c)$ has ? because it is the dot product of ?.

a)

b)

c)

d)

e)

f)

asked 2021-05-29

Find the vector and parametric equations for the line segment connecting P to Q.

P(0, - 1, 1), Q(1/2, 1/3, 1/4)

asked 2021-05-17

Find the scalar and vector projections of b onto a.

$a=(4,7,-4),b=(3,-1,1)$

asked 2022-07-25

If $A=-{a}_{x}+6{a}_{y}+5{a}_{z}andB={a}_{x}+2{a}_{y}+3{a}_{x}$, find (a) the scalar projection of A on B

asked 2022-09-25

If I use the rule of vector triple product, it becomes:

$\overrightarrow{b}\times \overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{a}(|\overrightarrow{b}{|}^{2})-\overrightarrow{b}(\overrightarrow{b}\cdot \overrightarrow{a})$

which is generally non-zero, but suppose I use properties of cross product:

$\overrightarrow{a}\times \overrightarrow{b}=-\overrightarrow{b}\times \overrightarrow{a}$

Hence,

$\overrightarrow{b}\times \overrightarrow{a}\times \overrightarrow{b}=-\overrightarrow{a}\times \overrightarrow{b}\times \overrightarrow{b}=\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{b})=0$

What did I do wrong?

$\overrightarrow{b}\times \overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{a}(|\overrightarrow{b}{|}^{2})-\overrightarrow{b}(\overrightarrow{b}\cdot \overrightarrow{a})$

which is generally non-zero, but suppose I use properties of cross product:

$\overrightarrow{a}\times \overrightarrow{b}=-\overrightarrow{b}\times \overrightarrow{a}$

Hence,

$\overrightarrow{b}\times \overrightarrow{a}\times \overrightarrow{b}=-\overrightarrow{a}\times \overrightarrow{b}\times \overrightarrow{b}=\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{b})=0$

What did I do wrong?

asked 2022-10-23

I have the vector $\overrightarrow{c}$ that is:

$\overrightarrow{c}=\frac{\sum _{i=1}^{n}{m}_{i}{\overrightarrow{r}}_{i}}{\sum _{i=1}^{n}{m}_{i}}$

where ${\overrightarrow{r}}_{i}$ is a vector and ${m}_{i}$ is a scalar

I need to proof the folowwing equality for any vector $\overrightarrow{r}$

$\sum _{i=1}^{n}{m}_{i}|\overrightarrow{r}-{\overrightarrow{r}}_{i}{|}^{2}=\sum _{i=1}^{n}{m}_{1}|{\overrightarrow{r}}_{i}-\overrightarrow{c}{|}^{2}+m|\overrightarrow{r}-\overrightarrow{c}{|}^{2}$

and i known that $m=\sum _{i=1}^{n}{m}_{i}$

I try to replace the vector $\overrightarrow{c}$ in the equality but but i get confused with the vector algebra.

$\overrightarrow{c}=\frac{\sum _{i=1}^{n}{m}_{i}{\overrightarrow{r}}_{i}}{\sum _{i=1}^{n}{m}_{i}}$

where ${\overrightarrow{r}}_{i}$ is a vector and ${m}_{i}$ is a scalar

I need to proof the folowwing equality for any vector $\overrightarrow{r}$

$\sum _{i=1}^{n}{m}_{i}|\overrightarrow{r}-{\overrightarrow{r}}_{i}{|}^{2}=\sum _{i=1}^{n}{m}_{1}|{\overrightarrow{r}}_{i}-\overrightarrow{c}{|}^{2}+m|\overrightarrow{r}-\overrightarrow{c}{|}^{2}$

and i known that $m=\sum _{i=1}^{n}{m}_{i}$

I try to replace the vector $\overrightarrow{c}$ in the equality but but i get confused with the vector algebra.