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Find the rank and the nullity of the following linear map from $\mathbb{R}^{4}$ to $\mathbb{R}^{3}$

$\left(x,y,z,t\right)\rightarrow(x-t,z-y,x-2y+2z-t)$

I understand how to find the rank and nullity, since the nullity is the dimension of the null space and the rank is the dimension of the column space. But I am having difficulty putting it into a matrix to solve.

And can someone also explain why we can take a linear map and essentially say that it is the "same" structurally as a the corresponding matrix. How do isomorphisms link into this idea ?

From this how would you know if it is onto ?

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You can write the transformation as $$T(\mathbf{x})=\mathbf{A}\mathbf{x}=\begin{bmatrix}1&0&0&-1\\0&-1&1&0\\1&-2&2&-1\end{bmatrix}\begin{bmatrix}x\\y\\z\\t\end{bmatrix}$$ where $\mathbf{A}$ is the transformation matrix. This matrix acts on a given vector $\mathbf{x}$ to give a transformed vector. If you carry out the matrix multiplication, you end up with $$\begin{bmatrix}1&0&0&-1\\0&-1&1&0\\1&-2&2&-1\end{bmatrix}\begin{bmatrix}x\\y\\z\\t\end{bmatrix}=\begin{bmatrix}x-t\\-y+z\\x-2y+2z-t\end{bmatrix}$$

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  • $\begingroup$ Can we not do it the other way around ? So have the columns as rows ? $\endgroup$ – Tom Otto May 13 '16 at 20:48
  • $\begingroup$ You can, each person tends to make a choice of whether there a row guy or a column guy. Each could me mapped to the other "isomorphically" which means that no Information is lost from the mapping. $\endgroup$ – shai horowitz May 13 '16 at 20:50
  • $\begingroup$ Can someone expand on how isomorphically these two objects are the same ? $\endgroup$ – Tom Otto May 13 '16 at 20:52
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There is a systematic way to construct a matrix representation of a linear transformation once ordered bases for the domain and range are chosen, however, for maps between $\mathbb{R}^n$ and $\mathbb{R}^m$ it is easy to figure out the matrix by inspection, if you understand how matrix multiplication works.

You want a matrix that will multiply on the left times a $4\times 1$ column vector and will give back a $3 \times 1$ column vector, so you know that the matrix must be a $3 \times 4$ matrix.

The first row of the matrix times the vector $\begin{bmatrix}x\\y\\z\\t\end{bmatrix}$ must equal the first component of the output vector, i.e. $x-t$. Similarly for rows 2 and 3. Thus it is easy to construct the desired matrix as $$ \begin{bmatrix} 1&0&0&-1\\0&-1&1&0\\1&-2&2&-1\end{bmatrix} $$

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You are given that the map is linear. So, this map (call it $T$) is a linear transformation. By definition, this means $T$ preserves sums and scalar products, i.e. if $T$ maps $a$ to $a'$ and $b$ to $b'$, then it maps $ca$ (for any scalar $c$) to $ca'$ and it maps $a+b$ to $a'+b'$.

Once we pick a basis for each of the two vector spaces $\mathbb{R}^4$ and $\mathbb{R}^3$, it can be proved that there is a 1-1 correspondence between linear transformations from $\mathbb{R}^4$ to $\mathbb{R}^3$ and $3 \times 4$ matrices over the real field. This statement requires a proof and can be shown. The matrices are represented with respect to bases for the two spaces. The idea behind the proof is that that every vector in the domain is a linear combination of the basis elements. So, if we know the image of the basis elements under the linear transformation $T$, we also know the image of every vector in the domain. The image of each basis element (of the domain) is a linear combination of basis elements in the codomain. So, if we know the coefficients of these linear combinations (i.e. how $T$ maps each basis element of the domain into a linear combination of the basis elements of the codomain), then we know $T$ entirely. These coefficients can be conveniently written in an 2D array (aka matrix) notation. Thus, we can represent linear transformations by matrices.

Also, if $A$ and $A'$ are matrices representing the linear transformations $T$ and $T'$, respectively, then we want the matrix representing the linear transformation $T T'$ to be the product $AA'$. Using this rule, we can derive the formula for multiplication of two matrices $A$ and $A'$ (you'll get the usual formula you've seen before, except that we now derived it).

Whether you take $3 \times 4$ or $4 \times 3$ matrices is up to you. Different authors/texts do it differently. The image of $x$ under $A$ can be written as $xA$ (where $x$ is a $1 \times 4$ row vector and $A$ is a $4 \times 3$ matrix) or can be written as $Ax$ (where $A$ is a $3 \times 4$ matrix and $x$ is a $4 \times 1$ column vector). In the former case, the image of the 4 basis elements of the domain form the rows of $A$, while in the latter convention the image of the 4 basis elements of the domain are put as the columns of $A$.

We usually take the basis vectors to be the unit vectors $e_i$ (which have a 1 in the $i$th coordinate and 0 in the remaining coordinates) because then the coefficients of the linear combination are obtained immediately to be the components of the vectors. If we do this, the $3 \times 4$ matrix $A$ can be written out. Observe that you are given that the linear transformation from $\mathbb{R}^4$ to $\mathbb{R}^3$ takes the basis element $e_1=(1,0,0,0)$ to $(1,0,1)$ (just substitute $x=1$ and substitute 0 for the other variables). So $(1,0,1)$ is the first column of your matrix. Similarly for the other three columns, and we get a $3 \times 4$ matrix whose columns are the image of the basis elements of the domain. (You can obtain this matrix more quickly by filling in the matrix row-wise using the given map. For example, put $(1,0,0,-1)$ as the first row since the formula given is $x-t$.)

We get a matrix whose third row is equal to the first row plus twice the second row. So, the rank of the matrix (which is defined to be the maximum number of linearly independent rows, and which can be shown to equal also the maximum number of linearly independent columns) is 2. Using the theorem that the rank and nullity add up to the dimension of domain 4, we get that the nullity is 2.

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