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Give the rank of the matrix

$$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Is the corresponding linear mapping injective, surjective, bijective?

Answer: the rank is three. Thus, the corresponding linear mapping is neither injective, nor surjective or bijective.

It is clear that the matrix has a rank of 3, since there are only three linearly independent columns in it. However, where do we get the properties of linear mapping and where is the mapping defined anyway? (don't see any corresponding notations) I'm rather new to this, so any readings are of great value. Thanks in advance.

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    $\begingroup$ Normally the mapping induced by a matrix is done so through matrix-vector multiplication. If $A$ is a matrix then the corresponding matrix mapping is given by $\mathbf{x}\mapsto A\mathbf{x}$. $\endgroup$ Commented Feb 28, 2013 at 17:27
  • $\begingroup$ I'm a little confused. You know that the matrix does not have full rank, and so it is not injective or surjective, but you don't know "where we get the properties"? You seem to have determined them just fine... $\endgroup$ Commented Feb 28, 2013 at 17:30
  • $\begingroup$ @rschwieb The confusion might be in linking properties of a matrix to the notions of injectivity/surjectivity as described when talking about functions from some set to another. These concepts do not seem linked to many students at first glance. $\endgroup$ Commented Feb 28, 2013 at 17:36
  • $\begingroup$ @Arkamis Yes, I think you're right that he meant "how are they connected" and not "where are the properties from." $\endgroup$ Commented Feb 28, 2013 at 17:43
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    $\begingroup$ Injective means that the nullspace is $\{0\}$, ie nullity equals $0$. Surjective means the range is the whole of $\mathbb{R}^4$, ie rank equals $4$. Bijective means injective+surjective. For a linear map between two finite dimensional spaces of same dimension, the rank-nullity theorem shows bijective $\Leftrightarrow$ injective $\Leftrightarrow$ surjective. $\endgroup$ Commented Mar 1, 2013 at 15:31

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An $n\times m$ matrix is interpreted as an $F$ linear transformation from $F^n$ to $F^m$ by multiplying row vectors from $F^n$ on their right side by this matrix. The result is a row vector in $F^m$. (Alternatively this can all be done with column vectors mutliplied on the left by this matrix, and then the map would be from $F^m$ to $F^n$.)

You should be able to see that it is indeed possible to find a vector which is sent to zero by this matrix, proving it is not injective.

If you know the so-called "rank-nullity theorem," it should connect this quite well with the injective and surjective conditions. It tells you how the rank of the matrix (which is the dimension of the image of the transformation) is connected with the dimension of the kernel and the dimension of the codomain (the space the function is going to).

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  • $\begingroup$ Based on the explanation the only vector which is sent to zero is the [0 0 0 0] vector. I don't see any problem here, cause if we take all the rows of the matrix (A) and multiply them by the matrix, we'll end up with the same rows. Which makes it a one to one mapping, thus bijective. rank A + nullity A = 3+1 = 4. So what I'm probably missing is what is our V and what is our W in T:V->W mapping. Can you please explain? $\endgroup$ Commented Mar 1, 2013 at 14:32
  • $\begingroup$ Hint: What is the image of $[0,0,1,0]$? $\endgroup$ Commented Mar 1, 2013 at 14:44
  • $\begingroup$ Yeah, ok, it mapps to the same [0, 0, 0, 0]. But from what I understood in explanation above it's not in the set of all row vectors. Which means that the problem is something like next: F:R^4->R^4:x->Ax. Then yes, two vectors are mapped to the same vector, which means that transformation is not injective (thus not bijective). I can imagine that there's the same vector [0, 0, 1, 0] which is in R^4 but is in no way mapped to it. Then transformation is also not surjective. Is that correct? $\endgroup$ Commented Mar 1, 2013 at 15:16
  • $\begingroup$ Right in both cases! If you look hard at what matrix mutlplication does here, it says that the image is exactly rowspace of the matrix. Thus, anything not in the rowspace is unreachable by the map! $\endgroup$ Commented Mar 1, 2013 at 15:20
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I guess the mapping is defined from/to $\mathbb{R}^3 \to \mathbb{R}^3$ if this is from a first course in linear algebra. I don't know how far you have come in your linear algebra course, but there is a theorem which says that a $n \times n$ matrix gives an injective linear map if and only if it gives a surjective map if and only if the matrix has $n$ linearly independent columns if and only if the matrix has a determinant different from $0$. This theorem doesn't necessarily hold if the matrix is not square i.e it is not an $n \times n$ matrix. You also have the famous rank theorem which states that for a linear operator $T$ between finite dimensional vector spaces you have that $rank(T) + dim \; ker(T) = dim \text{(dimension of the space the operator is defined on)}$.

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Here's one way to look at it.

Consider a differentiable function $f$ mapping $\Bbb R^n$ to $\Bbb R^m$.

As with single-variable functions, the derivative of $f$ is a linearization of $f$ around some point $\bf{x_0}$. The derivative of $f$ in this case is a matrix. So we write

$$f(\mathbf{x}+\mathbf{h}) = f(\mathbf{x})+f'(\mathbf{x})\mathbf{h} + o(\mathbf{h}^2)$$

One consequence of the locally linear nature of differentiability is that there is a neighborhood around $\mathbf{x_0}$ such that $f'$ is invertible. This is the core idea behind the Inverse Function Theorem.

So, if the function has a local inverse, then its derivative is invertible in this neighborhood. An invertible function must be bijective -- injective and onto -- and hence so must this linear mapping.

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