Timeline for Deducing properties of a transformation from its matrix
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Mar 1, 2013 at 15:31 | comment | added | Julien | 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. | |
| Mar 1, 2013 at 15:30 | vote | accept | Denys S. | ||
| Mar 1, 2013 at 15:07 | comment | added | Denys S. | @julien, oh, ok, sorry, but that is clear (given with the explanation), what I can't understand is how one can determine whether the transformation is injective, surjective or bijective. | |
| Mar 1, 2013 at 14:39 | comment | added | Julien | Let $\{e_1,e_2,e_3,e_4\}$ be the canonical basis of $\mathbb{R}^4$ with respect to which your matrix is taken. Then a basis of the nullspace is $\{e_3\}$ so the nullity is $1$. By the rank-nullity theorem, it follows that the rank is $4-1=3$. If you don't want to use this theorem, it is easily seen that a basis of the image is $\{e_1,e_2,e_4\}$. So the rank is $3$. | |
| Mar 1, 2013 at 14:35 | comment | added | Denys S. | @julien, how will it help me? | |
| Feb 28, 2013 at 18:13 | comment | added | Julien | It is almost trivial to determine a basis of the range and the nullspace. You should try to do so. | |
| Feb 28, 2013 at 18:06 | history | edited | Jim | CC BY-SA 3.0 | added 48 characters in body |
| Feb 28, 2013 at 17:43 | comment | added | rschwieb | @Arkamis Yes, I think you're right that he meant "how are they connected" and not "where are the properties from." | |
| Feb 28, 2013 at 17:36 | comment | added | Emily | @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. | |
| Feb 28, 2013 at 17:31 | answer | added | Oliver E. Anderson | timeline score: 2 | |
| Feb 28, 2013 at 17:30 | comment | added | rschwieb | 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... | |
| Feb 28, 2013 at 17:28 | answer | added | Emily | timeline score: 0 | |
| Feb 28, 2013 at 17:27 | comment | added | EuYu | 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 $\mathbf{x}\mapsto A\mathbf{x}$. | |
| Feb 28, 2013 at 17:26 | answer | added | rschwieb | timeline score: 2 | |
| Feb 28, 2013 at 17:24 | history | edited | rschwieb | CC BY-SA 3.0 | replaced title with a useful one. |
| Feb 28, 2013 at 17:22 | history | asked | Denys S. | CC BY-SA 3.0 |