# Search results

1. ### Basis of eigenvectors

Right the basis for the kernel is the span of $(1,1,1)$. Yes eigenvectors are linearly independent so they do span the range thanks!
2. ### Basis of eigenvectors

So the basis for the range of $T$ are the other two eigenvectors.
3. ### Basis of eigenvectors

Ok so the kernel of $T$ is $(x,y,z)$ such that $T(x,y,z)=0$ & this only occurs when we have $(1,1,1)$ so I guess that is the basis for the kernel right???
4. ### Basis of eigenvectors

Okay so I found the eigenvalues to be $\lambda = 0,-1,2$ with corresponding eigenvectors $v = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$. Not sure what to do next. Thanks!!!
5. ### Showing a matrix A is diagonalisable

Show that $\{\mathbf{v}_1\}$ is linearly independent. Simple enough lets consider $$c_1\mathbf{v}_1 = \mathbf{0}.$$ Our goal is to show that $c_1 = 0$. By the definition of eigenvalues and eigenvectors we have $A\mathbf{v}_1= \lambda_1\mathbf{v}_1$. Lets multiply the above equation and...

Thanks!
7. ### Finding $(g\circ f)'(6)$

The solution is 3: It's just $(g\circ f)'(6) = (-1,-2)\cdot (4,-3) = (-1\times 4)+((-2)\times (-3)) = -4+6 = 2$ using the multi-variate chain rule and the dot product. Is this correct and if not how do I go about doing it? Thanks!
8. ### Multivariate calculus problem: Calculating the gradient vector

Okay then I’m lost how do we then justify whether to use the chain rule???
9. ### Multivariate calculus problem: Calculating the gradient vector

Okay (1) and (2) are done. So for (3), assuming $t > 0$, $f\circ \mathbf{r} = \ln{(e^{\sin{(t)}})}^2+\ln{(e^{\sin{(t)}})}^2 = \sin^2{(t)}+\cos^2{(t)} = 1$ so the derivative is $0$.
10. ### Prove that the set T:={x∈Rn:Ax∈S} is a subspace of Rn.

Yeah that certainly doesn't make sense!!! 1. Suppose $\mathbf{0}\in T$, $A(\mathbf{0})\in S$ which is non-empty. 2. Suppose $x_1,x_2\in T$. We then have that $A(x_1+x_2) = A(x_1)+A(x_2)\in S$, i.e. $x_1+x_2\in T$. 3. Suppose $\mathbf{x}\in T$, with $\lambda\in \mathbb{R}$...
11. ### Prove that the set T:={x∈Rn:Ax∈S} is a subspace of Rn.

Thanks!!! 1.Since $\mathbf{0}\in T$, $A(\mathbf{0})\in S$ which is non-empty. 2. Suppose $x_1,x_2\in T$. Then there exists vectors $x_1,x_2\in S$ such that we have $A(x_1)$ and $A(x_2)$. We then have that $x_1+x_2\in S$ and $A(x_1+x_2) = A(x_1)+A(x_2)$, i.e. $x_1+x_2\in T$. 3...
12. ### Multivariate calculus problem: Calculating the gradient vector

The derivative of $\mathbf{r}$ at each point of $(0,1)$???
13. ### Multivariate calculus problem: Calculating the gradient vector

1. We find the partial derivatives of $f$ with respect to $x$ and $y$ to get $f_x = \frac{2\ln{(x)}}{x}$ and $f_y = \frac{2\ln{(y)}}{y}.$ This makes the gradient vector \nabla{f} = \begin{bmatrix} f_x \\ f_y \end{bmatrix} = \begin{bmatrix} \frac{2\ln{(x)}}{x} \\ \frac{2\ln{(y)}}{y}...

16. ### Area between 2 curves

Hold up. I can just calculate $\int_{0.5}^{1} (2x-1)^4 \; dx=0.1$. Then calculate $\int_{7/8}^{1} 8x-7 \; dx=0.0625$. Now take the difference to get $\frac{3}{80}=0.0375$.
17. ### Area between 2 curves

So that means I solve $\int_{0.5}^{1} (2x-1)^4 -8x + 7 \; dx$ in which case I get $0.6$???
18. ### Area between 2 curves

Oops no it isn’t that will be $x=1$.
19. ### Area between 2 curves

I’m thinking it’s where the tangent function cuts the $x$-axis which is when $8x-7=0$, or $x=\frac{7}{8}$.
20. ### Area between 2 curves

Yes it’s $x=0.5$.
21. ### Minimum length of wire

1. Homework Statement A wire pattern is inserted into a $10$cm square by making a horizontal line in the middle of the square (not all the way across and with length $x$) and connecting the ends of this line to the closest two corners. What is the minimum value of $x$? 2. Homework...