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Ker v. California

From Wikipedia, the free encyclopedia

Ker v. California
Argued December 11, 1962
Decided June 10, 1963
Full case nameDiane Ker, et. ux. v. California
Citations374 U.S. 23 (more)
83 S. Ct. 1623; 10 L. Ed. 2d 726; 1963 U.S. LEXIS 2473; 24 Ohio Op. 2d 201
Case history
PriorCert. to the District Court of Appeal of California, Second Appellate District
Holding
The Fourth Amendment’s prohibition on unreasonable search and seizure and the exclusionary rule for evidence obtained from unreasonable search and seizure apply to the states through the Fourteenth Amendment.
Court membership
Chief Justice
Earl Warren
Associate Justices
Hugo Black · William O. Douglas
Tom C. Clark · John M. Harlan II
William J. Brennan Jr. · Potter Stewart
Byron White · Arthur Goldberg
Case opinions
MajorityClark, joined by Black, Stewart, White
ConcurrenceHarlan (in judgment)
Concur/dissentBrennan, joined by Warren, Douglas, Goldberg
Laws applied
U.S. Const. amends. IV, XIV

Ker v. California, 374 U.S. 23 (1963), was a case before the United States Supreme Court, which incorporated the Fourth Amendment's protections against illegal search and seizure. The case was decided on June 10, 1963, by a vote of 5–4.

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  • Linear transformations | Matrix transformations | Linear Algebra | Khan Academy

Transcription

You now know what a transformation is, so let's introduce a special kind of transformation called a linear transformation. It only makes sense that we have something called a linear transformation because we're studying linear algebra. We already had linear combinations so we might as well have a linear transformation. And a linear transformation, by definition, is a transformation-- which we know is just a function. We could say it's from the set rn to rm -- It might be obvious in the next video why I'm being a little bit particular about that, although they are just arbitrary letters -- where the following two things have to be true. So something is a linear transformation if and only if the following thing is true. Let's say that we have two vectors. Say vector a and let's say vector b, are both members of rn. So they're both in our domain. So then this is a linear transformation if and only if I take the transformation of the sum of our two vectors. If I add them up first, that's equivalent to taking the transformation of each of the vectors and then summing them. That's my first condition for this to be a linear transformation. And the second one is, if I take the transformation of any scaled up version of a vector -- so let me just multiply vector a times some scalar or some real number c . If this is a linear transformation then this should be equal to c times the transformation of a. That seems pretty straightforward. Let's see if we can apply these rules to figure out if some actual transformations are linear or not. So let me define a transformation. Let's say that I have the transformation T. Part of my definition I'm going to tell you, it maps from r2 to r2. So if you give it a 2-tuple, right? Its domain is 2-tuple. So you give it an x1 and an x2 let's say it maps to, so this will be equal to, or it's associated with x1 plus x2. And then let's just say it's 3 times x1 is the second tuple. Or we could have written this more in vector form. This is kind of our tuple form. We could have written it -- and it's good to see all the different notations that you might encounter -- you could write it a transformation of some vector x, where the vector looks like this, x1, x2. Let me put a bracket there. It equals some new vector, x1 plus x2. And then the second component of the new vector would be 3x1. That's a completely legitimate way to express our transformation. And a third way, which I never see, but to me it kind of captures the essence of what a transformation is. It's just a mapping or it's just a function. We could say that the transformation is a mapping from any vector in r2 that looks like this: x1, x2, to-- and I'll do this notation-- a vector that looks like this. x1 plus x2 and then 3x1. All of these statements are equivalent. But our whole point of writing this is to figure out whether T is linearly independent. Sorry, not linearly independent. Whether it's a linear transformation. I was so obsessed with linear independence for so many videos, it's hard to get it out of my brain in this one. Whether it's a linear transformation. So let's test our two conditions. I have them up here. So let's take T of, let's say I have to vectors a and b. They're members of r2. So let me write it. A is equal to a1, a2, and b is equal to b1, b2. Sorry that's not a vector. I have to make sure that those are scalars. These are the components of a vector. And b2. So what is a1 plus b? Sorry, what is vector a plus vector b? Brain's malfunctioning. All right. Well, you just add up their components. This is the definition of vector addition. So it's a1 plus b1. Add up the first components. And the second components is just the sum of each of the vector's second compnents. a2 plus b2. Nothing new here. But what is the transformation of this vector? So the transformation of vector a plus vector b, we could write it like this. That would be the same thing as the transformation of this vector, which is just a1 plus b1 and a2 plus b2. Which we know it equals a vector. It equals this vector. Or what we do is for the first component here, we add up the two components on this side. So the first component here is going to be these two guys added up. So it's a1 plus a2 plus b1 plus b2. And then the second component by our transformation or function definition is just 3 times the first component in our domain, I guess you could say. So it's 3 times the first one. So it's going to be 3 times this first guy. So it's 3a1 plus 3b1. Fair enough. Now what is the transformation individually of a and b? So the transformation of a is equal to the transformation of a -- let me write it this way -- is equal to the transformation of a1 a2 in brackets. That's another way of writing vector a. And what is that equal to? That's our definition of our transformation right up here, so this is going to be equal to the vector a1 plus a2 and then 3 times a1. It just comes straight out of the definition. I essentially just replaced an x with a's. By the same argument, what is the transformation of our vector b? Well, it's just going to be the same thing with the a's replaced by the b's. So the transformation of our vector b is going to be -- b is just b1 b2 -- so it's going to be b1 plus b2. And then the second component in the transformation will be 3 times b1. Now, what is the transformation of vector a plus the transformation of vector b? Well, it's this vector plus that vector. And what is that equal to? Well, this is just pure vector addition so we just add up their components. So it's a1 plus a2 plus b1 plus b2. That's just that component plus that component. The second component is 3a1 and we're going to add it to that second component. So it's 3a1 plus 3b1. Now, we just showed you that if I take the transformations separately of each of the vectors and then add them up, I get the exact same thing as if I took the vectors and added them up first and then took the transformation. So we've met our first criteria. That the transformation of the sum of the vectors is the same thing as the sum of the transformations. Now let's see if this works with a random scalar. So we know what the transformation of a looks like. What does ca look like, first of all? I guess that's a good place to start. c times our vector a is going to be equal to c times a1. And then c times a2. That's our definition of scalar multiplication time's a vector. So what's our transformation -- let me go to a new color. What is our -- let me do a color I haven't used in a long time, white. What is our transformation of ca going to be? Well, that's the same thing as our transformation of ca1, ca2 which is equal to a new vector, where the first term -- let's go to our definition -- is you sum the first and second components. And then the second term is 3 times the first component. So our first term you sum them. So it's going to be ca1 plus ca2. And then our second term is 3 times our first term, so it's 3ca1. Now, what is this equal to? This is the same thing. We can view it as factoring out the c. This the same thing as c times the vector a1 plus a2. And then the second component is 3a1. But this thing right here, we already saw. This is the same thing as the transformation of a. So just like that, you see that the transformation of c times our vector a, for any vector a in r2 -- anything in r2 can be represented this way -- is the same thing as c times the transformation of a. So we've met our second condition, that when you when you -- well I just stated it, so I don't have to restate it. So we meet both conditions, which tells us that this is a linear transformation. And you might be thinking, OK, Sal, fair enough. How do I know that all transformations aren't linear transformations? Show me something that won't work. And here I'll do a very simple example. Let me define my transformation. Well, I'll do it from r2 to r2 just to kind of compare the two. I could have done it from r to r if wanted a simpler example. But I'm going to define my transformation. Let's say, my transformation of the vector x1, x2. Let's say it is equal to x1 squared and then 0, just like that. Let me see if this is a linear transformation. So the first question is, what's my transformation of a vector a? So my transformation of a vector a-- where a is just the same a that I did before-- it would look like this. It would look like a1 squared and then a 0. Now, what would be my transformation if I took c times a? Well, this is the same thing as c times a1 and c times a2. And by our transformation definition -- sorry, the transformation of c times this thing right here, because I'm taking the transformation on both sides. And by our transformation definition this will just be equal to a new vector that would be in our codomain, where the first term is just the first term of our input squared. So it's ca1 squared. And the second term is 0. What is this equal to? Let me switch colors. This is equal to c squared a1 squared and this is equal to 0. Now, if we can assume that c does not equal 0, this would be equal to what? Actually, it doesn't even matter. We don't even have to make that assumption. So this is the same thing. This is equal to c squared times the vector a1 squared 0. Which is equal to what? This expression right here is a transformation of a. So this is equal to c squared times the transformation of a. Let me do it in the same color. So what I've just showed you is, if I take the transformation of a vector being multiplied by a scalar quantity first, that that's equal to -- for this T, for this transformation that I've defined right here -- c squared times the transformation of a. And clearly this statement right here, or this choice of transformation, conflicts with this requirement for a linear transformation. If I have a c here I should see a c here. But in our case, I have a c here and I have a c squared here. So clearly this negates that statement. So this is not a linear transformation. And just to get a gut feel if you're just looking at something, whether it's going to be a linear transformation or not, if the transformation just involves linear combinations of the different components of the inputs, you're probably dealing with a linear transformation. If you start seeing things where the components start getting multiplied by each other or you start seeing squares or exponents, you're probably not dealing with a linear transformation. And then there's some functions that might be in a bit of a grey area, but it tends to be just linear combinations are going to lead to a linear transformation. But hopefully that gives you a good sense of things. And this leads up to what I think is one of the neatest outcomes, in the next video.

Prior history

George Douglas and Diane Ker (a married couple) were convicted of possession of marijuana in Southern California. The two were arrested after officers from the Los Angeles County Sheriff's Department saw George Ker meeting another person who was suspected of selling illegal drugs. Although lighting conditions and distance prevented the officers observing this meeting from seeing any exchange of money or drugs between Ker and the other man, they believed Ker was part of a drug-trading ring. After losing their prime suspect, the officers went to the Ker's apartment and entered without consent or a warrant using a pass key supplied by the building manager. An officer observed a “brick-shaped package of green leafy substance” on the kitchen table and arrested both Kers. A subsequent warrantless search of the apartment and the Ker's car found more packaged and loose marijuana and marijuana seeds, all of which was used as evidence against the Kers.

After conviction in state Superior Court, both the California District Court of Appeals and the California Supreme Court upheld the conviction, ruling that the evidence was not seized in the course of an unlawful search.

Case

The Court had decided two years earlier in Mapp v. Ohio that evidence seized in the course of an illegal search was inadmissible in a criminal trial in a state court. The Court extended that holding in this case, addressing the standard for deciding what are the fruits of an illegal search in state criminal trials. Clark's opinion addressed “the specific question as to whether Mapp requires the exclusion of evidence in this case which the California District Court of Appeal has held to be lawfully seized.” Unlike the previous case, where the search was clearly unreasonable, the District Court had found that the seizure of the drugs in the Kers’ apartment was allowed as being incident to an otherwise lawful arrest. The Supreme Court granted certiorari to give lower courts guidance on decide when evidence is the fruit of an unlawful search or seizure.

The Court declared that the standards of reasonableness are the same under the Fourth and Fourteenth Amendments applying in Federal and State courts. Clark interpreted the Court's precedents in determining if the search of the Kers’ apartment and car were reasonable.

Effects of the decision

The Kers themselves did not, however, personally benefit from this decision. The Supreme Court analyzed the evidence presented and agreed with the California courts that the seizure was incident to a lawful arrest. The Sheriff's officers had probable cause to make a warrantless arrest, and the most prominent evidence (the brick of marijuana) was in plain sight. Clark also dismissed the other reasonableness objections raised by the Kers. Brennan's opinion diverged from Clark's on this point, saying that the search was not reasonable under existing precedent.

Subsequent history

Ker has been cited by subsequent decisions both for the holding of Fourth Amendment incorporation and for allowing warrantless search and seizure with probable cause or to prevent destruction of contraband. For example, Michigan v. Tyler, 436 U.S. 499 (1978), and Wilson v. Arkansas, 514 U.S. 927 (1995).

See also

Further reading

  • Blakey, G. Robert (1964). "The Rule of Announcement and Unlawful Entry: Miller v. United States and Ker v. California". University of Pennsylvania Law Review. 112 (4): 499–562. doi:10.2307/3310634. JSTOR 3310634.

External links

This page was last edited on 13 September 2023, at 02:28
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