Mathematics, 05.05.2020 23:09, savyblue1724707

Let T be the linear transformation whose standard matrix is Aequals[Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column negative 2 3rd Column 3 2nd Row 1st Column negative 1 2nd Column 3 3rd Column negative 4 3rd Row 1st Column negative 5 2nd Column negative 2 3rd Column 9 EndMatrix ]. Determine whether the linear transformation T is one-to-one and whether it maps set of real numbers R cubed onto set of real numbers R cubed.

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answered: asdfghhk9805

Yes the linear transformation is one to one and it maps a set of real numbers R cubed onto set of real numbers R cubed

Step-by-step explanation:

To show that the linear transformation is one to one, we must show that the columns are linearly independent columns (i.e it must have the trivial solution)

$A = \left[\begin{array}{cccc}1&-2&3&0\\-1&3&-4&0\\-5&-2&9&0\end{array}\right]$

For the Augumented matrix A and applying Row Reduction Formula we have that

= $\left[\begin{array}{cccc}1&-2&3&0\\0&1&-1&0\\0&0&-18&0\end{array}\right]$

we can deductively see that $x_{3} = 0, x_{2} = 0, x_{1} = 0$

Since the columns are linearly independent, Hence it is a one-to-one transformation and it maps a set of real numbers R cubed onto set of real numbers R cubed

answered: ash10136

Step-by-step explanation:

The matrix is $\left[\begin{matrix}1 & -2 & 3\\ -1 & 3 & -4 \\ -5&-2 & 9\end{matrix}\right]$

Since this matrix represents a transformation, for us to check if it maps $\mathbb{R}^3$ onto

There is a theorem as a follows: A linear transformation is one to one if and only if the kernel of it's matrix representation is the vector 0. Recall that given a matrix A the kernel is defined as

KerA = $\{x | Ax = 0\}$

which is equivalent to solve the homogeneous system, and finding that the only solution is the vector 0. We will solve the system by gauss-jordan reduction.

If we apply gauss -jordan reduction to the matrix A, we get (the calculations of this matrix are beyond the scope of this answer)

$\begin{matrix}1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1\end{matrix}$

If we translate this to a equations system we get

x+z =0, y-z=0, z=0

We have that x=-z, y=z and z=0. Which shows that the only solution is the vector (0,0,0).

Hence, T is a one- to-one transformation. To check that T is a onto transformation, we only need to check that the reduced form of the matrix has the same number of pivots as the dimension of the codomain (in this case, R^3). Since $\mathbb{R}^3$ is a 3-dimensional vector space, the reduced matrix must have 3 pivots. So, since the matrix is 3x3 matrix, each column must have a pivot.

One can identify a pivot if it's a non zero entry whose entries that are on the same row and at the left, and all the entries of the same column that are under the entry are all 0.

We can check that in this case each column has the pivot 1 at columns 1, 2, and 3. So, since the reduced form has 3 pivots, T maps $\mathbb{R}^3$ onto

answered: Guest

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answered: Guest

answer: 2 answers

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